Predicting What Comes Next: Exploring Sequences and Progressions (Aage Kya Aayega? Sequences aur Progressions ki Khoj): Class 9 Maths Chapter 8 Hinglish mein

Ganita Manjari Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

95 min
1 simulation
41 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

Sequence ko shabdon mein aur t_n notation se describe karo, aur finite aur infinite sequences mein fark pehchano.
Explicit rule se koi bhi term nikaalo, aur position ke liye solve karke tay karo ki diya gaya number term hai ya nahi.
Recursive rule se terms banao, un rules ke saath bhi jo pichhle do terms use karte hain, jaise Virahānka–Fibonacci sequence.
Arithmetic progression pehchano, uska nth term a + (n − 1)d nikaalo, aur samjhao ki uske points ek straight line par kyun padte hain.
Terms ki jodiyan bana kar 1 + 2 + … + n = n(n + 1)/2 nikaalo, aur ise consecutive numbers aur simple APs ke sums ke liye use karo.
Geometric progression pehchano, uska nth term ar^(n − 1) nikaalo, aur compare karo ki APs aur GPs kaise badhte hain aur graph par kaise dikhte hain.
Sierpiński triangle jaise fractals ke pieces aur area ginne ke liye GPs use karo.

NCERT Mathematics Class 9, Chapter 8, pp. 174–196 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

8.1 Sequences ka Introduction

Maan lo kisi mahine ka pehla Sunday 3 tareekh ko padta hai. Calendar khole bina tum us mahine ke saare Sundays likh sakte ho: 3, 10, 17, 24 aur 31. Tumne ek rule notice kiya, 7 jodo, aur usse aage ka andaaza lagaya. Yeh chapter aisi hi numbers ki lists ke baare mein hai, unke andar chhupe rules ke baare mein, aur is baare mein ki ek achha rule hamein list mein bahut aage tak kaise jump karne deta hai.

Aisi kai lists tum pichhli classes se jaante ho. Aage padhne se pehle neeche wala box try karo.

Inmein se har list ek sequence (अनुक्रम) hai: ek tay order mein likhe gaye numbers, pehla, doosra, teesra aur aage. Sequence banane wale numbers uske terms kehlaate hain. Order matter karta hai. List 4, 1, 9 mein wahi numbers hain jo 1, 4, 9 mein hain, lekin yeh ek alag sequence hai. End mein teen dots ka matlab hai ki list hamesha chalti rehti hai.

Definition

Sequence

Ek tay order mein numbers ki list. Yeh numbers uske terms hain: pehla term, doosra term, aur aage.

Example2, 4, 6, 8, … even numbers ka sequence hai; iska teesra term 6 hai.

Kuch sequences ruk jaate hain. Cinema hall ki ek row ke seat numbers, 1 se 24, 24 terms ka ek finite sequence banate hain. Countdown 10, 9, 8, …, 1, 0 bhi finite hai, jismein 11 terms hain. Lekin 5 ke multiples kabhi khatam nahi hote, isliye 5, 10, 15, 20, … ek infinite sequence hai.

Table 8.1Chaar jaane-pehchaane sequences aur unke terms ke beech ke gaps

Sequence	Pehle paanch terms	Terms ke beech gaps	Ise dekhne ka ek aur tareeka
natural numbers	1, 2, 3, 4, 5	1, 1, 1, 1	1-1 karke aage gino
odd numbers	1, 3, 5, 7, 9	2, 2, 2, 2	2-2 karke aage gino
triangular numbers	1, 3, 6, 10, 15	2, 3, 4, 5	$1 + 2 + \dots + n$
square numbers	1, 4, 9, 16, 25	3, 5, 7, 9	$n \times n$

Naam “triangular” kahan se aaya? Ek plate par laddoos ko 1, 2, 3, … ki rows mein jamao, jaise Fig. 8.1 mein, to woh ek triangle ban jaate hain. Har nayi neeche wali row apne upar wali row se ek laddoo lambi hoti hai, isliye Table 8.1 mein gaps har baar ek se badhte hain: 2, 3, 4, 5, … Row-wise ginne se yeh bhi dikhta hai ki har triangular number counting numbers ka running total hai, jaise $10 = 1 + 2 + 3 + 4$ .

6/6Row-wise gino: $15 = 1 + 2 + 3 + 4 + 5$ .

Fig. 8.1Har nayi neeche wali row upar wali row se ek laddoo lambi hai.NCERT Fig. 8.1 · p. 175

Ek plate par gol laddoos ka pyramid hai, aur right mein ek panel total, use banane wala sum aur ab tak ka sequence dikhata hai. Step 1 mein ek akela laddoo hai: 1. Step 2 stack ko upar khiskata hai aur neeche 2 ki row jodta hai, jisse 3 = 1 + 2 banta hai. Steps 3, 4 aur 5 neeche 3, 4 aur 5 laddoos ki rows jodte hain, jisse 6, 10 aur 15 bante hain; har baar sabse nayi row blue mein dikhti hai. Neeche sequence 1, 3, 6, 10, 15 banta jaata hai, aur padosi numbers ke beech jumps +2, +3, +4, +5 dikhte hain. Aakhri step har row par uski ginti, 1 se 5, likhta hai aur 15 = 1 + 2 + 3 + 4 + 5 dikhata hai.

Squares bhi tiles se isi tarah step-by-step banaye ja sakte hain. Fig. 8.2 mein ek tile $2 \times 2$ square ban jaati hai jab 3 tiles ka ek L-shape uske around lag jaata hai, phir 5 ke L se $3 \times 3$ square banta hai, aur aage isi tarah. Har L-shape mein agle odd number jitni tiles hoti hain. Isliye pehle $n$ odd numbers ko jodne par hamesha $n^{2}$ milta hai, aur padosi square numbers ke beech ke gaps odd numbers 3, 5, 7, 9, … hote hain.

5/59 tiles ka L $5 \times 5 = 25$ deta hai.

Fig. 8.2Har L-shaped border mein agle odd number jitni tiles hoti hain, aur woh ek square ko agle square mein badal deta hai.NCERT Fig. 8.2 · p. 175

Square tiles bottom-left corner mein rakhi hain, aur right mein ek panel hai. Step 1 mein ek tile hai: 1 = 1 × 1. Step 2 top aur right ke saath 3 tiles ka L-shaped border naye colour mein jodta hai, jisse 1 + 3 = 4 = 2 × 2 banta hai. Steps 3, 4 aur 5 mein 5, 7 aur 9 tiles ke L-shaped borders judte hain, har ek apne colour mein, aur 9, 16 aur 25 tiles ke squares bante hain. Panel ab tak jode gaye odd numbers ko unke borders ke colour mein dikhata hai, aur total ko ek square ki tarah: 1 + 3 + 5 + 7 + 9 = 25 = 5 × 5.

Try karke dekhoNCERT Exercise · p. 176

Padhne ki streak

Meera ek reading streak shuru karti hai. Woh day 1 par 3 pages padhti hai, day 2 par 5 pages aur day 3 par 7 pages: har din pichhle din se 2 pages zyada.

Day 5, 6, 7 aur 8 par woh kitne pages padhti hai?
Ek doosri list banao: har din ke end tak usne total kitne pages padh liye. Yeh 3, 8, 15, … se shuru hoti hai. Ise day 10 tak aage badhao.
Apni doosri list ke har number mein 1 jodo. Tumne kya notice kiya? Kya tum ise Fig. 8.2 ki madad se samjha sakte ho?

Terms ko naam dena

Sequence ko lockers ki ek row samjho jin par 1, 2, 3, … number likhe hain, aur har locker mein ek number rakha hai. “Square numbers ke locker 4 mein rakha number” kehne ke bajaay hum $t_{4}$ likhte hain. Neeche likha chhota 4 subscript kehlaata hai: yeh locker number hai, yaani position. Poora symbol $t_{4}$ wahan rakhi cheez ko batata hai, yaani value. To square numbers ke liye $t_{4} = 16$ , jabki $t_{1} = 1$ aur $t_{2} = 4$ . General mein, $t_{n}$ position $n$ wala term hai (Fig. 8.3).

Fig. 8.3

t

ke neeche wala chhota number position hai; box mein likha number value hai:

t_{4} = 16

Ek row mein chhe boxes pehle chhe square numbers rakhte hain: 1, 4, 9, 16, 25 aur 36. Har box ke upar ek numbered blue circle uski position, 1 se 6, batata hai, aur har box ke neeche uska naam likha hai: 1 ke neeche t sub 1, 4 ke neeche t sub 2, aur isi tarah 36 ke neeche t sub 6 tak. Chautha box shaded hai aur uske around dashed line ka ghera hai, aur neeche likha text kehta hai ki position 4 par value 16 hai, yaani t sub 4 barabar 16.

Letter $t$ sirf ek aadat hai. Jab ek problem mein do sequences hon, to unhe alag letters do, jaise ek ke liye $s_{n}$ aur doosre ke liye $u_{n}$ , taaki woh mix na hon. Locker numbers hamesha counting numbers 1, 2, 3, … hote hain: 1st term aur 2nd term hota hai, lekin 2.5th term nahi hota. Haan, locker mein rakha number kuch bhi ho sakta hai. Neeche wale sequences Table 8.1 wale sequences se bilkul alag tarah behave karte hain.

$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots$ : fractions jo dheere-dheere 1 ki taraf badhte hain lekin kabhi 1 tak pahunchte nahi.
$- 12, - 9, - 6, - 3, 0, 3, \dots$ : negative terms jo ek baar mein 3 upar chadhte hain.
$1, - 2, 3, - 4, 5, \dots$ : terms jo positive aur negative ke beech jhoolte rehte hain.
$7, 7, 7, 7, \dots$ : ek constant sequence, jismein har term same hai.
$1, 2, 1, 2, 1, 2, \dots$ : ek pattern jo baar-baar repeat hota hai.

Quick check

4 mein se 1

1, 3, 6, 10, 15, 21, … ka agla term hai.

Answer dekho

Answer 28

Gaps 2, 3, 4, 5, 6 hain, isliye agla gap 7 hai aur

21 + 7 = 28

8.2 Sequence ka Explicit Rule

Sequence ka pattern jaanna kaam ka hai. $n$ ke terms mein $t_{n}$ ka formula jaanna aur bhi achha hai. Aise formula ko explicit rule kehte hain: position daalo, aur term bahar aa jaata hai.

Definition

Explicit rule

Ek formula jo sequence ka $n$ th term seedhe uski position $n$ se deta hai.

Example $t_{n} = n^{2}$ square numbers deta hai, isliye $t_{12} = 144$ .

Example 1NCERT Example 1 · p. 177

Woh numbers jo remainder 3 chhodte hain

Rule $t_{n} = 4 n - 1$ ek sequence describe karta hai. Iske pehle paanch terms aur 25th term likho. Iske saare terms mein kya common hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Rule se 250th term ek line mein mil jaata hai: $t_{250} = 4 \times 250 - 1 = 999$ . Baar-baar 4 jodne mein 249 additions lagenge. Rule ko ulta bhi chalaya ja sakta hai, value se wapas uski position tak. 3, 7, 11, … mein 199 kahan hai? Agar $199 = t_{n}$ , to $4 n - 1 = 199$ , isliye $4 n = 200$ aur $n = 50$ : 199 50th term hai.

Example 2NCERT Example 2 · p. 177

Highway par rest stops

Ek highway ke rest stops par 1, 2, 3, … number likhe hain, aur rest stop $n$ $(7 n + 3)$ km mark par hai, isliye pehle kuch stops 10, 17 aur 24 km marks par hain. Ek car 262 km mark par kharab ho jaati hai aur doosri 450 km mark par. Kya koi car bilkul rest stop par hai? Jo car stop par nahi hai, uske liye sabse paas wala stop kaun sa hai, aur woh kitni door hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3 $7 n + 3 = 450$ se $n \approx 63.9$ milta hai: 450 63rd aur 64th terms ke beech padta hai.

Fig. 8.4

7 n + 3

ke terms 7 ki doori par hain. Koi number tabhi term hai jab woh bilkul inmein se kisi ek par aa kar gire.

Upar wali row ek number line hai jis par 7n + 3 ke pehle chhe terms dots se mark hain: 10, 17, 24, 31, 38 aur 45, aur har do padosi terms ke beech +7 ka jump bana hai. Step 2 mein 262 ke paas ek zoomed number line aati hai, jismein unit tick marks hain aur terms 255, 262 aur 269 par 36th, 37th aur 38th likha hai; ek arrow 262 ko mark karta hai aur woh bilkul 37th term par girta hai, isliye answer hai yes, n = 37. Step 3 mein 450 ke paas doosri zoomed line terms 444 (63rd) aur 451 (64th) dikhati hai, aur unke beech ka gap shaded hai; 450 ka arrow gap ke andar girta hai, 451 se ek kam, isliye answer hai no, n lagbhag 63.9 hai.

Test hamesha ek hi tarah kaam karta hai. Rule ko us number ke barabar rakho aur $n$ ke liye solve karo. Agar $n$ counting number aata hai, to woh number ek term hai aur $n$ uski position hai. Agar $n$ fraction, zero ya negative aata hai, to woh number sequence mein nahi aata.

Squares ke liye rule bas $t_{n} = n^{2}$ hai. Primes ki kahani alag hai. Log do hazaar saal se zyada samay se unhe study kar rahe hain, aur koi simple formula $n$ th prime nahi deta. Har sequence ka saaf-suthra explicit rule nahi hota.

Quick check

4 mein se 1

t_{n} = 4 n - 1

use karke 60th term nikaalo.

Answer dekho

Answer 239

t_{60} = 4 \times 60 - 1 = 239

8.3 Sequence ka Recursive Rule

Aksar sequence ke baare mein sabse aasaan baat yeh hoti hai ki woh ek term se agle term tak kaise pahunchta hai. Savings plan kehta hai “pichhle hafte se ₹5 zyada daalo”, aur staircase plan kehta hai “har step neeche wale step se 5 cm ooncha hai”. Aise instructions ek-ek term karke sequence banate hain. Letters mein likhne par “6 se shuru karo aur 5 jodte raho” ban jaata hai $t_{1} = 6$ aur $n \geq 2$ ke liye $t_{n} = t_{n - 1} + 5$ , jahan $t_{n - 1}$ ka matlab hai $t_{n}$ se theek pehle wala term. Isse 6, 11, 16, 21, 26, … milta hai, wahi list jo explicit rule $t_{n} = 5 n + 1$ deta hai.

Is tarah ka rule recursive rule (पुनरावर्ती नियम) kehlaata hai. Akele yeh shuru nahi ho sakta: ise pehla term batana padta hai, kyunki pehle term se pehle koi term nahi hota jisse use banaya ja sake. Fig. 8.5 dono tarah ke rules ko compare karta hai.

Fig. 8.5Recursive rule ek term se doosre term tak chalta hai; explicit rule seedha tumhare chahe term par jump karta hai.

Upar wali row, jiska title walk hai, terms 6, 11, 16, 21, 26 aur 31 ko gol stepping stones ki tarah dikhati hai, har stone se agle tak +5 wala ek curved arrow hai, aur ek note kehta hai ki 6th term tak pahunchne mein 5 steps lagte hain: yeh recursive rule hai. Neeche wali row, jiska title jump hai, n = 40 wala ek box dikhati hai jo 5 × 40 + 1 wale box ki taraf point karta hai, aur woh answer 201 ki taraf, yaani 40th term, jo pehle ke 39 terms se guzre bina mil gaya.

Definition

Recursive rule

Aisa rule jo sequence ka har term usse pehle ke ek ya zyada terms se deta hai, saath mein starting term ya terms bhi batata hai.

Example $t_{1} = 6$ aur $t_{n} = t_{n - 1} + 5$ se 6, 11, 16, 21, … milte hain

Har tarah ke rule ki apni khoobi hai. Recursive rule aksar pehchanna aasaan hota hai, kyunki woh batata hai ki ek step se agle step tak kya hota hai. Door ke terms ke liye explicit rule jaldi kaam karta hai, kyunki use pehle ke kisi term ki zaroorat hi nahi hoti.

Example 3NCERT Example 3 · p. 178

Ek tank jo settle ho jaata hai

Subah 6 baje school ke water tank mein 200 litres paani hai. Har ghante tank ka aadha paani use ho jaata hai, aur phir ek pump 40 litres daalta hai. Maan lo $u_{1} = 200$ subah 6 baje ka amount hai, $u_{2}$ 7 baje ka amount, aur aage isi tarah, taaki $n \geq 2$ ke liye $u_{n} = \frac{1}{2} u_{n - 1} + 40$ . (a) Pehle chhe terms nikaalo. (b) Kya 100 ek term hai? (c) Kya tank mein kabhi bilkul 80 litres honge?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 178

Ek sequence jo bahut tezi se bhaagta hai

$s_{1} = 2$ se shuru karo. Baad ka har term usse pehle ke saare terms ke product se 1 zyada hai: $n \geq 2$ ke liye $s_{n} = s_{1} \times s_{2} \times \dots \times s_{n - 1} + 1$ . Pehle paanch terms likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4 wale sequence mein ek surprise chhupa hai. Iske terms ke reciprocals jodo: $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ , phir $\frac{5}{6} + \frac{1}{7} = \frac{41}{42}$ , phir $\frac{41}{42} + \frac{1}{43} = \frac{1805}{1806}$ . Har naya term total ko 1 ke aur paas le aata hai, phir bhi total kabhi poora 1 nahi banta.

Virahānka–Fibonacci sequence

Purane India ke kaviyon ke saamne ginti ki ek puzzle aayi jiske liye aisa rule chahiye jo do terms peeche dekhe. Sanskrit aur Prakrit poetry mein syllable ya to short hota hai, jo ek beat leta hai, ya long, jo do beats leta hai. $n$ beats ki line ko kitni alag rhythms bhar sakti hain? Fig. 8.6 2, 3 aur 4 beats ke liye unki list deta hai.

Har rhythm ya to short syllable par khatam hoti hai ya long par. Woh aakhri syllable hata do, to jo bachta hai woh ek beat chhoti ya do beat chhoti rhythm hai. Isliye $n$ beats ki rhythms ki ginti $n - 1$ beats ki ginti plus $n - 2$ beats ki ginti hai. 1 beat ke liye 1 rhythm aur 2 beats ke liye 2 rhythms se shuru karne par ginti 1, 2, 3, 5, 8, 13, 21, … chalti hai. Letter $V$ ke saath likhne par rule hai $V_{1} = 1$ , $V_{2} = 2$ aur 3 se aage har $n$ ke liye $V_{n} = V_{n - 2} + V_{n - 1}$ . Yahi Virahānka–Fibonacci sequence hai.

4/44 beats ki $3 + 2 = 5$ rhythms. Agla: $5 + 3 = 8$ .

Fig. 8.6Har rhythm short ya long syllable par khatam hoti hai, isliye har ginti apne se pehle ki do gintiyon ka sum hai.

Ek legend short syllable ko 1 beat wale chhote blue block aur long syllable ko 2 beats wale chaude orange block ki tarah dikhata hai. Step 1 mein 2 beats bharne wale saare patterns (short-short, long: 2 ways) aur 3 beats wale (short-short-short, short-long, long-short: 3 ways) ki list hai. Step 2 mein 4 beats ka column shuru hota hai, un teen patterns se jo short syllable par khatam hote hain; har ek 3-beat pattern hai jismein ek short joda gaya hai, aur joda gaya block black outline mein hai. Step 3 woh do patterns jodta hai jo long syllable par khatam hote hain, har ek 2-beat pattern mein ek long jodkar. Step 4 3 + 2 = 5 ways ginta hai aur running ginti 1, 2, 3, 5, 8, 13, … dikhata hai, jismein nayi ginti, 5, highlighted hai.

Timeline

Rhythms kisne gini?

7th century CE
Virahānka
Metre par likhi kitaab Vṛttajātisamuccaya mein rhythms ginne ka rule deta hai.
c. 1135 CE
Gopāla
Metres par ek commentary mein wahi ginti ka rule phir se samjhata hai.
c. 1150 CE
Hemachandra
Metre par apne granth mein is rule ko use karta hai.
1202 CE
Leonardo of Pisa (Fibonacci)
Apni kitaab Liber Abaci mein khargoshon ke badhne ki ek puzzle mein wahi numbers paata hai.

Yahi numbers maths mein aur nature mein baar-baar dikhte hain, jaise sunflower ke phool mein beejon ke spirals mein.

Quick check

4 mein se 1

Virahānka–Fibonacci sequence mein 8 ke baad ke teen terms 13, 21 aur 34 hain. Unke baad ke do terms aur hain.

Answer dekho

Answer (1) 55, (2) 89

21 + 34 = 55

aur phir

34 + 55 = 89

Practice · 7 questions

Practice 8.1: answers ke saath

1
Short answerEasyNCERT Exercise Set 8.1, Q1
Har sequence ke pehle paanch terms likho.
1. $t_{n} = 4 n - 9$
  Solution dikhao
  Answer $- 5, - 1, 3, 7, 11$
  $4 - 9 = - 5$ , $8 - 9 = - 1$ , $12 - 9 = 3$ , $16 - 9 = 7$ , $20 - 9 = 11$ .
2. $t_{n} = \frac{n + 3}{2}$
  Solution dikhao
  Answer $2, \frac{5}{2}, 3, \frac{7}{2}, 4$
  $\frac{4}{2} = 2$ , $\frac{5}{2}$ , $\frac{6}{2} = 3$ , $\frac{7}{2}$ , $\frac{8}{2} = 4$ : terms har baar $\frac{1}{2}$ badhte hain.
3. $t_{n} = (- 1)^{n} (2 n + 1)$
  Solution dikhao
  Answer $- 3, 5, - 7, 9, - 11$
  $(- 1)^{n}$ odd $n$ ke liye $- 1$ aur even $n$ ke liye $+ 1$ hai, isliye signs baari-baari badalte hain: $- 3$ , $5$ , $- 7$ , $9$ , $- 11$ .
2
Short answerEasyNCERT Exercise Set 8.1, Q2
Sequence $t_{n} = 6 n + 5$ ke liye neeche diye terms nikaalo.
1. 12th term
  Answer dekho
  Answer 77
  $6 \times 12 + 5 = 77$ .
2. 20th term
  Answer dekho
  Answer 125
  $6 \times 20 + 5 = 125$ .
Short answerMediumNCERT Exercise Set 8.1, Q3
Priya kehti hai ki 100 aur 150 dono sequence 9, 16, 23, 30, … mein aate hain. Kya woh sahi hai? Pehle ek explicit rule dhoondho.
Solution dikhao
AnswerSirf aadhi sahi: 100 14th term hai, lekin 150 term nahi hai.
Terms 7-7 badhte hain aur $t_{1} = 9 = 7 (1) + 2$ , isliye $t_{n} = 7 n + 2$ .
$7 n + 2 = 100$ se $7 n = 98$ aur $n = 14$ milta hai, jo counting number hai. Isliye $100 = t_{14}$ .
$7 n + 2 = 150$ se $7 n = 148$ aur $n = 21 \frac{1}{7}$ milta hai, jo counting number nahi hai. Isliye 150 term nahi hai: woh $t_{21} = 149$ aur $t_{22} = 156$ ke beech mein hai.
Short answerEasyNCERT Exercise Set 8.1, Q4
Rahul 5 se shuru karke 8-8 ki ginti aage karta hai: 5, 13, 21, 29, … Jab woh 461 bolta hai, tab tak usne kitne numbers bol diye?
Answer dekho
Answer 58
$t_{n} = 5 + 8 (n - 1) = 8 n - 3$ . Phir $8 n - 3 = 461$ se $8 n = 464$ milta hai, isliye $n = 58$ : 461 uska bola gaya 58th number hai.
5
Short answerMediumNCERT Exercise Set 8.1, Q5
Freezer se nikali gayi ek test tube $- 10$ °C par hai. Agle aadhe ghante tak uska temperature har 5 minute mein 3 °C badhta hai. Tum turant thermometer padhte ho aur phir har 5 minute mein, isliye pehli reading $t_{1} = - 10$ hai.
1. Readings ke liye recursive rule likho aur pehli paanch readings ki list banao.
  Solution dikhao
  Answer $t_{1} = - 10$ , $t_{n} = t_{n - 1} + 3$ ; $- 10, - 7, - 4, - 1, 2$
  Har reading pichhli se 3 zyada hai, isliye $t_{1} = - 10$ aur $n \geq 2$ ke liye $t_{n} = t_{n - 1} + 3$ . Readings $- 10$ , $- 7$ , $- 4$ , $- 1$ aur 2 (°C mein) hain.
2. Kaun si reading 8 °C hai?
  Answer dekho
  Answer 7
  $t_{n} = - 10 + 3 (n - 1) = 3 n - 13$ , aur $3 n - 13 = 8$ se $n = 7$ milta hai: 7th reading, jo pehli reading ke 30 minute baad li gayi.
3. Kya koi reading bilkul 0 °C dikhati hai? Samjhao.
  Solution dikhao
  AnswerNahi.
  $3 n - 13 = 0$ se $n = 4 \frac{1}{3}$ milta hai, jo counting number nahi hai. Temperature 4th reading ( $- 1$ °C) aur 5th reading (2 °C) ke beech 0 °C ko paar karta hai.
6
Short answerMediumNCERT Exercise Set 8.1, Q6
$A_{1} = A_{2} = A_{3} = 1$ se shuru karo. Uske baad har term apne theek pehle ke teen terms ka sum hai: $n \geq 4$ ke liye $A_{n} = A_{n - 1} + A_{n - 2} + A_{n - 3}$ .
1. $A_{4}, A_{5}, A_{6}, A_{7}$ aur $A_{8}$ nikaalo.
  Solution dikhao
  Answer3, 5, 9, 17, 31
  $A_{4} = 1 + 1 + 1 = 3$ , $A_{5} = 3 + 1 + 1 = 5$ , $A_{6} = 5 + 3 + 1 = 9$ , $A_{7} = 9 + 5 + 3 = 17$ , $A_{8} = 17 + 9 + 5 = 31$ .
2. Ab tak ka har term odd hai. Samjhao ki baad ka har term bhi odd kyun hoga.
  Solution dikhao
  AnswerTeen odd numbers ka sum odd hota hai.
  Odd + odd even hota hai, aur even + odd odd hota hai, isliye teen odd numbers ka sum odd hai. Pehle teen terms odd hain, isliye $A_{4}$ odd hai. Phir $A_{5}$ bhi teen odd terms ka sum hai, aur yahi step hamesha repeat hota rehta hai.
Quick answerEasy
Kaun sa explicit rule sequence 9, 14, 19, 24, … deta hai?
$t_{n} = 9 n + 5$ $t_{n} = 4 n + 5$ $t_{n} = 5 n + 4$ $t_{n} = 5 n + 9$
Answer dekho
Answer (C) $t_{n} = 5 n + 4$
$5 (1) + 4 = 9$ aur $5 (2) + 4 = 14$ . Rule $5 n + 9$ 14 ko pehla term bana deta.

8.4 Arithmetic Progressions

Sabse orderly sequences woh hain jo har step par bilkul same amount se badalte hain. Dhyaan se dekhoge to yeh har jagah milenge: row mein seat numbers, ruler par marks, ek jaisi plates ke dher ki height.

Kai Indian kitchens mein steel ke tumblers ek ke andar ek rakhne ke liye bane hote hain. Har tumbler 12 cm lamba hai, aur ek tumbler ko doosre ke andar rakhne par woh neeche wale ke rim se 2 cm upar nikla rehta hai. Fig. 8.7 ek stack ko bante hue dikhata hai.

5/5Chaar glasses: pehle ke 12 cm plus teen extra 2 cm ke tukde.

Fig. 8.7Har extra tumbler stack mein wahi 2 cm jodta hai.NCERT Fig. 8.3 · p. 180

Left mein ek centimetre ruler khada hai aur uske paas ek steel tumbler; right mein ek table glasses ki ginti aur stack ki height batati hai. Step 1 mein ek tumbler hai, 12 cm lamba, aur uske rim se ruler tak ek dashed line hai. Steps 2, 3 aur 4 mein doosra, teesra aur chautha tumbler stack mein daala jaata hai; har ek neeche wale ke andar baithta hai aur 2 cm bahar nikla rehta hai, isliye heights 14 cm, 16 cm aur 18 cm hoti hain aur table row-row karke badhti hai. Aakhri step chaar glasses ki height 12 + 2 + 2 + 2 = 12 + 3 × 2 likhta hai.

Heights 12, 14, 16, 18, … centimetres hain. Socho ki $n$ tumblers ka stack kis cheez se banta hai: ek poora tumbler, 12 cm lamba, aur uske upar 2 cm ke $n - 1$ rims, pehle ke baad har tumbler ka ek. Isliye uski height $t_{n} = 12 + (n - 1) \times 2 = 2 n + 10$ centimetres hai. Paanch tumblers 20 cm oonche hain aur aath 26 cm, aur kyunki $2 n + 10 = 100$ se $n = 45$ milta hai, 45 tumblers ka stack bilkul 1 metre ooncha hai. $n - 1$ par dhyaan do: pehla tumbler 2 cm ka rim nahi jodta.

12, 14, 16, 18, … mein har term apne se pehle wale term se bilkul 2 zyada hai. Is property wale sequences itne common hain ki unka apna naam hai.

Definition

Arithmetic progression (AP)

Aisa sequence jo har step par ek fixed amount $d$ se upar ya neeche jaata hai, taaki saare $n \geq 2$ ke liye $t_{n} - t_{n - 1} = d$ . Is constant step $d$ ko hum common difference kehte hain.

Example12, 14, 16, 18, … ek AP hai jiska first term 12 aur common difference 2 hai.

Arithmetic progression ko hum short mein AP (समांतर श्रेढ़ी) kehte hain. Har baar joda jaane wala fixed amount common difference $d$ hai. Ise nikaalne ke liye kisi bhi term ko uske agle term se subtract karo: $d = t_{2} - t_{1} = t_{3} - t_{2} = \dots$ . Common difference negative bhi ho sakta hai. 25, 18, 11, 4, −3, … mein har term pichhle se 7 kam hai, isliye $d = - 7$ .

Har AP do numbers se fix hota hai: uska first term $a$ aur uska common difference $d$ . Uske terms $a, a + d, a + 2 d, a + 3 d, \dots$ hain. Position 1 se position $n$ tak jaane ke liye hum size $d$ ke $n - 1$ barabar hops lete hain, jaise Fig. 8.8 dikhata hai.

4/4 $t_{n} = a + (n - 1) d$ : first term mein $n - 1$ hops jodo.

Fig. 8.8

n

th term tak pahunchne ke liye first term se size

d

n - 1

barabar hops lagte hain.

Upar wali number line terms 12, 14, 16, 18 aur 20 ko blue dots se mark karti hai. Step 2 mein +2 likha ek green arrow har term se agle tak hop karta hai, aur ek note kehta hai ki 5th term pehle se 4 hops door hai: 12 + 4 × 2 = 20. Step 3 mein neeche ek doosri number line aati hai jis par terms 25, 18, 11, 4 aur −3 purple dots hain; −7 likhe dashed orange arrows har term se agle tak left ki taraf hop karte hain, aur ek note 5th term 25 + 4 × (−7) = −3 deta hai. Aakhri step general rule jodta hai: nth term first term plus size d ke n − 1 hops hai.

PropertyAP ka

n

th term

Agar kisi AP ka first term $a$ aur common difference $d$ hai, to uska $n$ th term $t_{n} = a + (n - 1) d$ hai.

AP ka recursive rule bhi bahut simple hai: $t_{1} = a$ aur $n \geq 2$ ke liye $t_{n} = t_{n - 1} + d$ . Tumblers ke liye $t_{1} = 12$ aur $t_{n} = t_{n - 1} + 2$ . Recursive rule ek step batata hai; explicit rule $t_{n} = a + (n - 1) d$ batata hai ki saare steps hamein kahan le jaate hain.

Example 5NCERT Example 5 · p. 183

Gullak bharna

Apne birthday par Kavya ek nayi gullak mein ₹120 daalti hai. Uske baad woh har Sunday ₹25 daalti hai. (a) 1, 2, 3 aur 4 Sundays ke baad gullak mein amount likho. (b) $n$ Sundays ke baad aur 8 Sundays ke baad amount nikaalo. (c) Kitne Sundays ke baad amount pehli baar ₹500 se upar jaata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Neeche diya Sequence builder tumhe apne khud ke APs banane deta hai aur har ek ko ek saath teen tarah dikhata hai: list ki tarah, changes aur ratios ki table ki tarah, aur graph par dots ki tarah. Tum isse door ke terms bhi poochh sakte ho aur test kar sakte ho ki koi number term hai ya nahi. Section 8.6 mein sequences ki ek doosri family ke liye hum iske paas wapas aayenge.

InteractiveSequence builder

Ek sequence builder. Buttons se rule ka type chuno: add d (ek AP), multiply by r (ek GP) ya aakhri do terms jodna. Steppers first term aur d set karte hain, ek menu r set karta hai, aur ek stepper aakhri-do-jodo rule ke liye second term set karta hai. Ek table pehle aath terms, pichhle term se change aur ratio ke saath dikhati hai, aur har column ke neeche batati hai ki woh same rehta hai ya nahi. Ek dot plot terms ko unki positions ke against dikhata hai, saath mein pehle do dots se guzarti ek optional dashed line hai jo us par pade har dot ko ghere mein le leti hai. Ek panel explicit aur recursive rules likhta hai, ek status line batati hai ki sequence AP hai ya GP, aur do tools kisi bhi position ka term nikaalte hain aur test karte hain ki koi number term hai ya nahi. Mystery mode ek chhupe AP ya GP ke pehle chaar terms hints ke saath dikhata hai, tumhe match karne ke liye.

Fig. 8.9Ek AP, GP ya “aakhri do jodo” wala sequence banao, phir uske changes, ratios aur graph compare karo.

Quick check

4 mein se 1

Inmein se kaun se sequences APs hain?

Saare sahi options chuno

Answer dekho

Answer

(A) odd numbers 1, 3, 5, 7, …
(C) 10, 17, 24, 31, …
(E) 35, 28, 21, 14, …

Odd numbers (

d = 2

), 10, 17, 24, … (

d = 7

) aur 35, 28, 21, … (

d = - 7

) ka difference constant hai. Triangular numbers ke gaps 2, 3, 4, … hain aur Virahānka–Fibonacci numbers ke gaps 1, 1, 2, 3, … hain, jo badalte rehte hain.

8.4.1 AP ko Visualise karna

Chalo tumbler heights ko graph par rakhte hain. Tumblers ki ginti ko $x$ aur stack ki height ko $y$ lo, aur Table 8.2 ke pairs plot karo.

Table 8.2Tumblers ke stack ki height

Tumblers ki ginti	1	2	3	4	5	$n$
Height (cm)	12	14	16	18	20	$2 n + 10$

4/4Ek step aur peeche jaane par 0 glasses par $12 - 2 = 10$ milta hai: $t_{n} = 2 n + 10$ .

Fig. 8.10Barabar steps aage chalne se barabar steps upar milte hain, isliye AP ke points ek line mein aate hain.NCERT Fig. 8.4 · p. 182

Ek grid mein aage glasses ki ginti 0 se 6 tak aur upar height 0 se 24 cm tak hai. Step 1 points (1, 12), (2, 14), (3, 16), (4, 18) aur (5, 20) ko blue dots ki tarah plot karta hai. Step 2 un sabse guzarti ek straight line kheenchta hai. Step 3 line ke neeche ek seedhi jodta hai: har point se 1 glass aage ka ek green step aur 2 cm upar ka ek purple step agle point tak pahunchata hai. Step 4 line ko dashed segment ki tarah peeche 0 glasses tak badhata hai, jahan woh height axis ko 10 cm par milti hai, jise ghere mein label kiya gaya hai.

Paanchon points ek straight line par hain, aur yeh koi coincidence nahi hai. Right mein ek step ka matlab hamesha ek tumbler zyada hai, aur ek tumbler zyada hamesha wahi 2 cm jodta hai, isliye graph par har step same amount upar chadhta hai. Line ko peeche badhao to woh height axis ko 10 par milti hai, jo value rule $2 n + 10$ $n = 0$ ke liye deta hai.

PropertyAP ke points ek line par hote hain

First term $a$ aur common difference $d$ wale kisi bhi AP ke liye, har point $(n, t_{n})$ straight line $y = d x + (a - d)$ par hota hai.

Isliye AP ka common difference uski line ka slope hai. Bada $d$ steep line deta hai, aur negative $d$ aisi line deta hai jo left se right girti hai. Neeche ke simulation mein tum points khud plot karte ho.

SimulationEk AP plot karo, phir ek GP

x = −1 se 7 aur y = −4 se 9 tak ka coordinate grid, jismein do fixed points hain, A (1, −3) par aur B (2, −1) par, jo ek AP ke pehle do terms hain. Grid par tap karne se, ya arrow keys se cursor chala kar Enter dabane se, ek naya point plot hota hai jo whole numbers par snap hota hai; har point label hota hai aur uske coordinates list mein aate hain. Undo aur Clear tumhare apne points hatate hain lekin A aur B rehne dete hain.

Fig. 8.11A aur B AP −3, −1, 1, 3, 5, … ke points

(1, t_{1})

aur

(2, t_{2})

hain. Aur terms plot karne ke liye grid par tap karo.NCERT Exercise · p. 182

Try karo

AP −3, −1, 1, 3, 5 ke 3rd, 4th aur 5th terms ko $x = 3, 4$ aur $5$ par plot karo. Kya paanchon points ek straight line par hain?
Har point se agle tak grid squares gino: kitne aage, aur kitne upar? Upar ke step ko AP ke common difference se compare karo.
Clear dabao (A aur B rehte hain; is baar unhe chhod do), phir doubling sequence 2, 4, 8 se $(1, 2)$ , $(2, 4)$ aur $(3, 8)$ plot karo. Kya yeh teen points ek line mein aate hain, ya mud jaate hain?

Quick check

3 mein se 1

AP 50, 46, 42, 38, … ke points

(n, t_{n})

ek straight line par hain jo left se right girti hai.

SahiGalat

Answer dekho

Answer Sahi

Har AP straight line par points deta hai. Yahan

d = - 4

, isliye right mein har step 4 neeche jaata hai.

8.5 Pehle n Natural Numbers ka Sum

Socho ek ladki ne apne 1st se 12th tak har birthday par sahi ginti ki candles wala cake kaata hai. Usne kul kitni candles bujhayi hain? Hamein $1 + 2 + 3 + \dots + 12$ chahiye. Ek-ek number jodna kaam karta hai, lekin ek zyada smart tareeka hai jo kahin lambe sums ke liye bhi utna hi achha chalta hai.

Numbers ko kaagaz ki ek patti par likho aur use beech se modo, jaise Fig. 8.12 mein. Modne se 12 1 ke upar aata hai, 11 2 ke upar, aur isi tarah 7 6 ke upar tak. Har pair ka sum 13 hai, aur 6 pairs hain, isliye total $6 \times 13 = 78$ candles hai.

4/413 ke 6 pairs: $6 \times 13 = 78$ .

Fig. 8.12Modne se sabse chhota number sabse bade ke saath pair banata hai: har pair ka sum same hai.

Upar baarah cells ki ek patti hai jis par 1 se 12 likha hai: left mein 1 se 6 blue mein, right mein 7 se 12 orange mein, aur 6 aur 7 ke beech ek dashed fold line hai. Step 2 mein right wala aadha left wale aadhe ke neeche mud jaata hai, isliye 12 1 ke neeche aata hai, 11 2 ke neeche, aur isi tarah 7 6 ke neeche tak. Step 3 mein do numbers ka har column ek green bracket se juda hai aur us par 13 likha hai. Step 4 mein text kehta hai ki 6 pairs hain, har ek 13, isliye 6 × 13 = 78, aur 1 + 2 + … + 12 = 78.

Jab numbers ki ginti even ho, tab modne se numbers ke pairs achhe se ban jaate hain. Thoda alag nazariya is idea ko har $n$ ke liye kaam karne laayak bana deta hai: list ko modne ki jagah use do baar likho, ek baar seedha aur ek baar ulta.

TheoremPehle

n

natural numbers ka sum

1 se $n$ tak ke natural numbers ka sum $\frac{n (n + 1)}{2}$ hai. Yaani $1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$ .

Fig. 8.13 wahi argument blocks se dikhata hai. 1, 2, 3, 4 aur 5 blocks ki rows wali seedhi, aur aadha ghumayi gayi uski ek copy, milkar bilkul 5 blocks oonchi aur 6 blocks chaudi rectangle bhar deti hain. Isliye do seedhiyon mein $5 \times 6 = 30$ blocks hain, aur ek seedhi mein 15.

4/4Saath mein: 6 ki 5 rows, isliye $2 S = 5 \times 6 = 30$ aur $S = 15$ .

Fig. 8.13Do barabar seedhiyan ek rectangle banati hain, isliye ek seedhi uska aadha hai.NCERT Fig. 8.5 · p. 184

Step 1 mein blue square blocks ki ek seedhi hai jismein 1, 2, 3, 4 aur 5 blocks ki rows hain, har row par uski ginti likhi hai. Step 2 uske right mein ek barabar orange seedhi dikhata hai. Step 3 orange seedhi ko aadha ghuma deta hai, taaki uski lambi row upar ho. Step 4 use blue seedhi ke paas ke gaps mein khiska deta hai: dono milkar 6 blocks chaudi aur 5 blocks oonchi rectangle banati hain, jis par 6 aur 5 likha hai, aur jismein 5 × 6 = 30 blocks hain, isliye har seedhi mein 15 hain.

Ab se $1 + 2 + \dots + n$ ka short naam $S_{n}$ hoga. Isliye $S_{n} = \frac{n (n + 1)}{2}$ ; jaise $S_{5} = 15$ aur $S_{12} = 78$ .

$1 + 2 + \dots + 15$ ke liye 15 columns hain, har ek ka sum 16, isliye $2 S = 15 \times 16 = 240$ aur $S = 120$ . $1 + 2 + \dots + 100$ ko ulta karne par 100 columns milte hain, har ek ka sum 101. Isliye $S_{100} = \frac{100 \times 101}{2} = 5050$ .

Example

Consecutive numbers ka ek block

$31 + 32 + 33 + \dots + 70$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Section 8.1 ke triangular numbers bilkul yahi sums hain: $n$ th triangular number $S_{n} = \frac{n (n + 1)}{2}$ hai. Isliye 12th triangular number 78 hai, yaani utni hi candles jitni humne gini thi.

Example

7 ke multiples jodna

7 se 98 tak 7 ke saare multiples ka sum nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

S_{40} = 1 + 2 + \dots + 40

nikaalo.

Answer dekho

Answer 820

S_{40} = \frac{40 \times 41}{2} = 820

Practice · 8 questions

Practice 8.2: answers ke saath

1
Short answerEasyNCERT Exercise Set 8.2, Q1
AP 7, 13, 19, 25, … ke liye neeche diye terms nikaalo.
1. 12th term
  Answer dekho
  Answer 73
  $7 + 11 \times 6 = 73$ .
2. 30th term
  Answer dekho
  Answer 181
  $7 + 29 \times 6 = 181$ .
2
Short answerMediumNCERT Exercise Set 8.2, Q2
Meena 45 se 4-4 karke ulti ginti karti hai: 45, 41, 37, 33, … Woh kehti hai, “Agar main chalti rahi, to bilkul 0 par pahunch jaaungi.”
1. Kya Meena sahi hai? Reason do.
  Solution dikhao
  AnswerNahi.
  $49 - 4 n = 0$ se $n = 12.25$ milta hai, jo counting number nahi hai. Woh $t_{12} = 1$ bolti hai aur phir $t_{13} = - 3$ , yaani 0 ko laangh jaati hai.
2. Kis position par woh pehli baar negative number bolti hai?
  Answer dekho
  Answer 13
  $49 - 4 n < 0$ ke liye $n > 12.25$ chahiye, isliye pehla negative term $t_{13} = 49 - 52 = - 3$ hai.
Short answerEasyNCERT Exercise Set 8.2, Q3
AP 14, 9, 4, −1, … ka $n$ th term nikaalo, iska recursive rule likho, aur iska 15th term nikaalo.
Solution dikhao
Answer $t_{n} = 19 - 5 n$ ; recursive rule $t_{1} = 14$ , $t_{n} = t_{n - 1} - 5$ ; $t_{15} = - 56$ .
Yahan $a = 14$ aur $d = 9 - 14 = - 5$ , isliye $t_{n} = 14 + (n - 1) (- 5) = 19 - 5 n$ . $n = 4$ check karo: $19 - 20 = - 1$ .
Har term pichhle se 5 kam hai: $t_{1} = 14$ aur $n \geq 2$ ke liye $t_{n} = t_{n - 1} - 5$ .
Explicit rule se $t_{15} = 19 - 5 \times 15 = - 56$ .
4
Short answerHardNCERT Exercise Set 8.2, Q4
Ek auditorium mein seats ki 30 rows hain. Har row mein apne aage wali row se utni hi extra seats hain. Row 6 mein 32 seats hain aur row 14 mein 48 seats.
1. Row 1 mein kitni seats hain?
  Answer dekho
  Answer 22
  Subtract karne par $8 d = 16$ , isliye $d = 2$ . Phir $a = 32 - 5 \times 2 = 22$ .
2. Sabse peechhe wali row, row 30, mein kitni seats hain?
  Answer dekho
  Answer 80
  $t_{30} = 22 + 29 \times 2 = 80$ .
3. Auditorium mein kul kitni seats hain?
  Answer dekho
  Answer 1530
  Āryabhaṭa ke rule se total $30 \times \frac{22 + 80}{2} = 30 \times 51 = 1530$ hai.
5
Short answerMediumNCERT Exercise Set 8.2, Q5
8 ke un three-digit multiples ko lo jo 200 se kam hain.
1. Woh kitne hain?
  Answer dekho
  Answer 12
  Woh $8 \times 13$ se $8 \times 24$ tak hain, yaani $24 - 13 + 1 = 12$ numbers.
2. Unka sum kya hai?
  Answer dekho
  Answer 1776
  $8 (13 + 14 + \dots + 24) = 8 (S_{24} - S_{12}) = 8 (300 - 78) = 1776$ . Āryabhaṭa ka rule bhi yahi deta hai: $12 \times \frac{104 + 192}{2} = 12 \times 148 = 1776$ .
Short answerMediumNCERT Exercise Set 8.2, Q6
Ek cycle factory apne pehle saal mein 1200 bicycles banati hai aur har saal production 150 bicycles badhati hai. Kis saal mein woh pehli baar 2800 se zyada bicycles banati hai?
Answer dekho
Answer 12
$(n - 1) \times 150 > 1600$ se $n - 1 > 10.67$ milta hai, isliye sabse chhota counting number $n$ jo kaam karta hai, 12 hai. Check karo: saal 11 mein 2700 bicycles bante hain, jo kaafi nahi, aur saal 12 mein 2850.
Short answerEasyNCERT Exercise Set 8.2, Q7
Mahine ki 1st tareekh ko Arjun ek jar mein ₹1 daalta hai, 2nd ko ₹2, 3rd ko ₹3, aur isi tarah aage. 24th din ke paise daalne ke baad jar mein kitne rupees hain?
rupees
Answer dekho
Answer 300 rupees
$S_{24} = \frac{24 \times 25}{2} = 300$ , isliye jar mein ₹300 hain.
Quick answerMedium
Sum $2 + 4 + 6 + \dots + 100$ barabar hai
2550505025005100
Answer dekho
Answer (A) 2550
Yeh $2 (1 + 2 + \dots + 50) = 2 \times 1275 = 2550$ hai. Value 5050 $1 + 2 + \dots + 100$ hai, jo odd numbers ko bhi jod deta hai.

8.6 Geometric Progressions

Ab ek aisa sequence jo bilkul alag tarah badhta hai. Kaagaz ki ek sheet ko aadha modo, phir se aadha, aur aise hi karte raho. Har fold ke baad sheet kholo aur creases se bane panels gino. Fig. 8.14 dikhata hai ki tumhe kya milta hai.

6/6Har fold panels ko 2 se aur ek panel ke size ko $\frac{1}{2}$ se multiply karta hai.

Fig. 8.14Har fold panels ki ginti double karta hai aur har panel ka size aadha karta hai.NCERT Fig. 8.6 · p. 186

Left mein ek rectangular sheet hai aur right mein ek table jiske columns folds, panels aur 1 panel hain. Step 1 poori sheet shaded dikhata hai: 0 folds, size 1 ka 1 panel. Step 2 beech mein upar se neeche ek dashed fold line jodta hai: 1 fold, 2 panels, ek panel shaded, size aadha. Steps 3, 4 aur 5 baari-baari se aadi, khadi aur phir aadi fold lines jodte hain, jisse 4, 8 aur 16 panels bante hain; shaded panel sheet ka ek-chauthai, ek-aathwan aur ek-solahwan hissa reh jaata hai, aur table row-row karke badhti hai. Aakhri step panels column par times 2 aur size column par times one half mark karta hai.

Panels ki gintiyan 2, 4, 8, 16, … hain. Fold ek fixed number of panels jodta nahi hai. Woh unhe double karta hai, kyunki har panel do mein bant jaata hai. Isliye $n$ folds ke baad $2 \times 2 \times \dots \times 2 = 2^{n}$ panels hote hain. Paanch folds se 32 panels milte hain. Nau folds se sirf 512 milte hain, lekin das folds se $2^{10} = 1024$ , yaani hazaar se zyada, aur bees folds se das lakh se zyada ( $2^{20} = 1 048 576$ ). Asal mein, haath se modi gayi normal sheet lagbhag saat folds ke baad itni moti ho jaati hai ki phir modi nahi jaati.

Yahan tumblers wali trick kaam nahi karti: padosiyon ko subtract karne par 2, 4, 8, … milte hain, jo barabar nahi hain. Iske bajaay divide karna kaam karta hai: $\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = 2$ , isliye har ginti apne se pehle wali ki bilkul double hai. Jo list har step par same multiplier se badhti ya ghatti hai, woh geometric progression hai.

Definition

Geometric progression (GP)

Non-zero numbers ka aisa sequence jismein kisi bhi term ko usse pehle wale term se divide karne par hamesha same number $r$ milta hai, taaki saare $n \geq 2$ ke liye $t_{n} = r \times t_{n - 1}$ . Is constant multiplier $r$ ko common ratio kehte hain.

Example2, 4, 8, 16, … ek GP hai jiska first term 2 aur common ratio 2 hai.

Geometric progression ko hum short mein GP (गुणोत्तर श्रेढ़ी) kehte hain. Fixed multiplier common ratio $r$ hai. Ise nikaalne ke liye kisi bhi term ko usse pehle wale term se divide karo: $r = \frac{t_{2}}{t_{1}} = \frac{t_{3}}{t_{2}} = \dots$ . First term $a$ wala GP $a, a r, a r^{2}, a r^{3}, \dots$ chalta hai.

PropertyGP ka

n

th term

Agar kisi GP ka first term $a$ aur common ratio $r$ hai, to uska $n$ th term $t_{n} = a r^{n - 1}$ hai.

GP ka recursive rule bhi utna hi chhota hai: $t_{1} = a$ aur $n \geq 2$ ke liye $t_{n} = r \times t_{n - 1}$ . Dono families ko saath rakh kar dekho. AP har step par same number jodta hai, jabki GP same number se multiply karta hai.

Example 6NCERT Example 6 · p. 187

10% se badhte paise

Ek bank har saal ke end mein savings account ke paison mein 10% jodta hai. Saal 1, 2, 3 aur 4 ki shuruaat mein Rohan ka balance ₹2000, ₹2200, ₹2420 aur ₹2662 hai. Dikhao ki yeh amounts ek GP banate hain, aur agar woh kuch na nikaale to saal 6 ki shuruaat mein uska balance nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 187

Har baar aadha

Ek ribbon ko aadha kaata jaata hai, phir ek aadhe ko aadha, aur isi tarah aage. Lengths, cm mein, 64, 32, 16, 8, … hain. Check karo ki yeh lengths GP banati hain, aur agli teen likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 187

Palatte signs

Do lists ek hi tarah shuru hoti hain: 2, −6, 18, −54, … aur 2, −6, 18, −30, … Inmein se kaun si GP hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 187

Fraction wala ratio

Dikhao ki 8, 12, 18, 27, … ek GP hai aur iska $n$ th term nikaalo. Iska 5th term kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

8.6.1 Fractals ke saath Maza

Sierpiński triangle, sabse mashhoor fractals mein se ek, ek hi instruction ko hamesha repeat karne se banta hai: har shaded triangle mein teeno sides ke midpoints jodo aur beech wala ulta triangle kaat do. Fig. 8.15 stage 0 par ek solid triangle se shuru hota hai jiski teeno sides barabar hain. Ek round kaatne se stage 1 milta hai, bache hue har tukde par doosra round stage 2 deta hai, aur rounds ko kabhi rukna nahi padta. Sierpiński triangle woh shape hai jiske paas yeh stages aur paas aate jaate hain.

5/5Triangles: har stage $\times 3$ . Area: har stage $\times \frac{3}{4}$ .

Fig. 8.15Har stage har shaded triangle ko teen chhote triangles se badal deta hai aur shaded area ka ek-chauthai hata deta hai.NCERT Fig. 8.7 · p. 188

Left mein ek bada shaded equilateral triangle hai; right mein ek panel stage, shaded triangles ki ginti aur shaded area batata hai. Step 1 stage 0 hai: ek triangle, area 1. Step 2 stage 1 hai: sides ke midpoints jodne se beech ka ek ulta triangle kat jaata hai, aur teen-chauthai area wale 3 shaded triangles bachte hain. Step 3 stage 2 hai: har shaded triangle ka beech wala hissa kat jaata hai, aur nau-solahwan area wale 9 triangles bachte hain. Step 4 stage 3 hai: 27 triangles, area sattais-chaunsathwan. Aakhri step ginti ke paas times 3 each stage aur area ke paas times three quarters each stage jodta hai.

Socho aur samjhoNCERT Think and Reflect · p. 189

Gino aur naapo

Aage padhne se pehle Fig. 8.15 se pehli row bharo: stages 0 se 3 tak shaded triangles gino, phir stages 4, 5 aur $n$ predict karo.

Doosri row ke liye stage 0 wale solid triangle ka area 1 square unit maano aur nikaalo ki kaatne ka har round kitna area chhodta hai.

Stage	0	1	2	3	4	5	$n$
Shaded triangles
Shaded area

Tumhari entries apne aap save hoti hain.

Ab area wali row ko dekho. Stages aage badhte jaane par shaded area kis taraf ja raha hai?

Socho ki ek cut ek shaded triangle ke saath kya karta hai: woh teen shaded triangles chhod deta hai, har ek aadha chauda. Isliye ginti jo bhi ho, agle stage mein usse teen guna hoti hai: $1 \to 3 \to 9 \to 27$ . $n$ stages ke baad humne bilkul $n$ baar 3 se multiply kiya hai, jisse $3^{n}$ triangles milte hain; stage 5 mein hi 243 hain. (Yahan hum stages ko 0 se number karte hain, isliye first term stage 0 ki ginti hai.)

Area ulti taraf jaata hai. Midpoint lines har shaded triangle ko chaar barabar tukdon mein baant deti hain, aur unmein se sirf teen shaded rehte hain. Isliye har stage shuruaat ke shaded area ka $\frac{3}{4}$ rakhta hai, aur $n$ stages ke baad area original ka ${(\frac{3}{4})}^{n}$ hai: $1, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \dots$

Table 8.3Sierpiński triangle se do GPs

Stage	0	1	2	3	4	5	$n$
Shaded triangles	1	3	9	27	81	243	$3^{n}$
Shaded area	1	$\frac{3}{4}$	$\frac{9}{16}$	$\frac{27}{64}$	$\frac{81}{256}$	$\frac{243}{1024}$	${(\frac{3}{4})}^{n}$

To ek fractal hamein ek saath do GPs deta hai, jinke common ratios 3 aur $\frac{3}{4}$ hain. Symbols mein, $t_{n} = 3^{n}$ jahan $t_{0} = 1$ aur $t_{n} = 3 t_{n - 1}$ , jabki $s_{n} = {(\frac{3}{4})}^{n}$ jahan $s_{0} = 1$ aur $s_{n} = \frac{3}{4} s_{n - 1}$ . Triangles ki ginti bina kisi limit ke badhti hai, phir bhi unka total area ghatta hi jaata hai. Stage 10 tak area lagbhag 0.06 hai, aur stage 20 tak 0.004 se kam: yeh 0 ke jitna chaho utna paas pahunch jaata hai.

Definition

Fractal

Aisi shape jo apni hi chhoti aur chhoti copies se bani ho, aksar ek simple step ko baar-baar repeat karke banayi jaati hai.

ExampleSierpiński triangle: har shaded tukda poori shape ki ek chhoti copy hai.

8.6.2 GP ko Visualise karna

GP ke terms graph par kaise dikhte hain? Table 8.4 ke paper-folding counts plot karo, aage folds ki ginti aur upar panels ki ginti.

Table 8.4Sheet ko baar-baar aadha modne se bane panels

Folds	0	1	2	3	4	5	6	$n$
Panels	1	2	4	8	16	32	64	$2^{n}$

4/4 $r = \frac{1}{2}$ ke saath points girte hain aur 0 ki taraf flat ho jaate hain.

Fig. 8.16Double karne se points kisi bhi straight line se door mud jaate hain; aadha karne se woh 0 ki taraf baith jaate hain.NCERT Fig. 8.9 · p. 191

Left graph mein aage folds 0 se 6 tak aur upar panels 0 se 64 tak hain. Step 1 (0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32) aur (6, 64) ko blue dots ki tarah plot karta hai, aur right mein ek list har point se agle tak ka change deti hai: +1, +2, +4, +8, +16, +32, har baar double. Step 2 pehle do points se ek dashed straight line kheenchta hai; 6 folds par woh sirf 7 tak pahunchti hai, 64 wale point se bahut neeche. Step 3 dots ko ek smooth curve se jodta hai jo aur zyada steep hokar upar mudta hai. Step 4 us list ki jagah ek panel ke size ka doosra graph dikhata hai, 1 se ek-chaunsathwan tak: uske green dots har baar aadhe hote hain aur curve 0 ki taraf flat hota jaata hai, lekin 0 tak pahunchta nahi.

Is baar points straight line par nahi hain. Pehle do points se guzarti line har fold par 1 panel chadhti hai, lekin asli gintiyan 1, phir 2, phir 4, phir 8 chadhti hain: chadhai khud har step par double hoti hai. Isliye points ek curve ke saath upar mudte hain jo aur steep aur steep hota jaata hai. Jis GP ka ratio 0 aur 1 ke beech ho, jaise ek panel ka size, woh ulti taraf mudta hai. Uske points pehle tezi se girte hain, phir level ho jaate hain, aur 0 ke paas aur paas aate jaate hain.

Dono families ke beech ka fark sabse saaf isi picture mein dikhta hai. AP ke points hamesha straight line par hote hain. GP ke points kabhi nahi, jab tak uska ratio 1 na ho aur har term same na ho.

Try karke dekhoNCERT Fig. 8.10 · p. 191

Wapas Sequence builder par

Sequence builder phir se kholo.

Multiply by r (GP) chuno, first term 2 aur $r = 2$ ke saath, yaani paper-folding panels. Line through the first two dots par tick karo. Ab kitne dots line par hain? Change column dekho: changes khud kya banate hain?
First term 1 aur $r = 3$ set karo, Sierpiński counts ki tarah. Phir $r = \frac{3}{4}$ set karo, unke areas ki tarah. Dono dot patterns describe karo.
Ab $r = - 2$ set karo. Terms ke signs ka kya hota hai, aur dots ka?

Example 10NCERT Example 10 · p. 192

Ganda paani filter karna

Ek purifier gande paani ko filter stages ki ek row se guzarta hai. Har stage apne tak pahunchi gandagi ka sirf 60% aage jaane deta hai, aur andar jaane wale paani mein 500 units dirt hai. (a) Pehle paanch stages mein se har ek ke baad aur $n$ stages ke baad bachi dirt nikaalo. (b) Purifier se nikle paani mein zyada se zyada 50 units dirt ho sakti hai. Kitne stages chahiye? (c) Ek behtar tarah ka stage sirf 40% dirt aage jaane deta hai. Aise kitne stages chahiye honge?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4/450 units ki limit ke saath 5 stages chahiye: 4 stages ke baad bhi 64.8 bachta hai.

Fig. 8.17Har stage apne tak pahunchi dirt ka 60% aage jaane deta hai.NCERT Fig. 8.11 · p. 193

Paani ke chhe jars ek row mein khade hain, chhote filters se hokar arrows se jude hue: ek jar filter se pehle ke paani ka aur ek-ek jar paanch filter stages mein se har ek ke baad ka. Har jar ke neeche jami dirt scale ke hisaab se bani hai, aur uska amount jar ke upar likha hai. Step 1 sirf pehla jar dikhata hai, 500 units dirt ke saath. Step 2 stages 1, 2 aur 3 ke baad ke jars jodta hai, jinmein 300, 180 aur 108 units hain, aur note kehta hai ki har stage dirt ko 0.6 se multiply karta hai. Step 3 stages 4 aur 5 ke baad ke jars jodta hai, 64.8 aur 38.9 units ke saath. Step 4 saare jars ke aar-paar 50 units ki limit par ek dashed line kheenchta hai: stage 4 ke baad ki dirt abhi bhi line ke upar tak hai aur us par too much likha hai, jabki stage 5 ke baad wala jar, green ghere mein aur OK mark kiya hua, pehla jar hai jo line ke neeche hai.

Kai asli quantities isi tarah ghatti hain: blood mein dawai, scooter ki resale value, dheemi padti goonj ki awaaz. Har ek time ke har step mein apna same fraction bachata hai, aur isse 0 aur 1 ke beech ratio wala GP banta hai.

Quick check

3 mein se 1

GP 1, 3, 9, 27, … ke points

(n, t_{n})

par kaun sa description fit hota hai?

woh ek curve par hain jo aur steep aur steep hota jaata haiwoh upar jaati straight line par hainwoh ek curve par hain jo 0 ki taraf level ho jaata haiwoh axis ke upar-neeche zigzag karte hain

Answer dekho

Answer (A) woh ek curve par hain jo aur steep aur steep hota jaata hai

Ratio 3, 1 se bada hai, isliye har chadhai pichhli ki 3 guna hai aur curve aur steep hota jaata hai. Straight line ke liye barabar chadhai chahiye, level hote curve ke liye 0 aur 1 ke beech ratio chahiye, aur zigzag ke liye negative ratio.

Neeche ka question 7 Fig. 8.18 wala fractal use karta hai. Har stage par har shaded square 9 barabar squares mein kaata jaata hai, aur sirf centre wala square aur sides ke beech wale chaar squares rakhe jaate hain.

4/4Ek baar aur dohrao.

Fig. 8.18Question 7 mein use hone wale cross fractal ke stages 0 se 3.NCERT Fig. 8.12 · p. 194

Ek bada shaded square dikhaya gaya hai aur right mein ek chhoti key hai: ek 3 by 3 grid jismein centre square aur har side ke beech ko chhoone wale chaar squares rakhe jaate hain, aur chaar corner squares hata diye jaate hain. Step 1 stage 0 hai, poora square. Step 2 stage 1 hai: chaar corner squares kaat diye jaate hain, aur plus ke shape ka cross bachta hai. Step 3 stage 2 hai: har rakhe gaye square par wahi cut lagta hai, isliye cross ki har bhuja ek chhota cross ban jaati hai. Step 4 stage 3 hai, jismein har rakhe gaye square par cut ek baar aur dohraya gaya hai.

Practice · 8 questions

Practice 8.3: answers ke saath

1
Short answerEasyNCERT Exercise Set 8.3, Q1
Ek GP mein $t_{4} = 54$ hai aur har term apne se pehle wale term ka 3 guna hai.
1. $t_{1}$ nikaalo.
  Answer dekho
  Answer 2
  4th term se 1st tak wapas jaane mein teen baar 3 se divide hota hai: $54 \div 27 = 2$ .
2. $t_{7}$ nikaalo.
  Answer dekho
  Answer 1458
  4th term se 7th tak aage jaane mein teen baar 3 se multiply hota hai: $54 \times 27 = 1458$ .
2
Short answerEasyNCERT Exercise Set 8.3, Q2
GP 4, 12, 36, … ke liye:
1. 8th term nikaalo.
  Answer dekho
  Answer 8748
  $t_{8} = 4 \times 3^{7} = 4 \times 2187 = 8748$ .
2. $n$ th term nikaalo.
  Solution dikhao
  Answer $t_{n} = 4 \times 3^{n - 1}$
  $t_{n} = a r^{n - 1} = 4 \times 3^{n - 1}$ .
3
Long answerHardNCERT Exercise Set 8.3, Q3
Tara kaagaz ki 1 sheet se shuru karti hai. Har round mein woh apne paas ke har tukde ko 3 tukdon mein kaatti hai aur phir 1 tukda de deti hai. Maan lo $t_{1} = 1$ kuch bhi kaatne se pehle tukdon ki ginti hai, $t_{2}$ round 1 ke baad ki ginti, aur aage isi tarah.
1. $t_{n}$ ke liye recursive rule likho aur $t_{2}$ se $t_{6}$ tak nikaalo.
  Solution dikhao
  Answer $t_{n} = 3 t_{n - 1} - 1$ ; 2, 5, 14, 41, 122
  Har tukde ko 3 mein kaatne se ginti teen guna hoti hai, aur ek de dene se 1 kam hota hai, isliye $t_{1} = 1$ aur $n \geq 2$ ke liye $t_{n} = 3 t_{n - 1} - 1$ . Phir $t_{2} = 2$ , $t_{3} = 5$ , $t_{4} = 14$ , $t_{5} = 41$ aur $t_{6} = 122$ .
2. Dikhao ki numbers $2 t_{n} - 1$ ek GP banate hain, aur iski madad se $t_{n}$ ka explicit rule likho.
  Solution dikhao
  Answer $2 t_{n} - 1 = 3^{n - 1}$ , isliye $t_{n} = \frac{3^{n - 1} + 1}{2}$ .
  Rule se, $2 t_{n} - 1 = 6 t_{n - 1} - 3 = 3 (2 t_{n - 1} - 1)$ . Isliye $2 t_{n} - 1$ ki har value pichhli ki 3 guna hai: 1, 3, 9, 27, 81, …, yaani first term 1 aur ratio 3 wala GP.
  Isliye $2 t_{n} - 1 = 3^{n - 1}$ aur $t_{n} = \frac{3^{n - 1} + 1}{2}$ . Check karo: $t_{6} = \frac{243 + 1}{2} = 122$ .
3. Sequence ka kaun sa term 3281 hai? Kitne rounds ke baad Tara ke paas itne tukde hote hain?
  Answer dekho
  Answer 9
  $\frac{3^{n - 1} + 1}{2} = 3281$ se $3^{n - 1} = 6561 = 3^{8}$ milta hai, isliye $n = 9$ : 9th term, jo 8 rounds ke baad aata hai, kyunki $t_{1}$ kuch bhi kaatne se pehle ke tukde ginta hai.
Short answerMediumNCERT Exercise Set 8.3, Q4
Aditi chessboard par doubling game khelti hai: square 1 par chawal ke 7 daane, square 2 par 14, square 3 par 28, aur aage isi tarah, har baar double. Kis square par 3584 daane hain? Square $n$ par daanon ki ginti ke liye explicit rule aur recursive rule likho.
Solution dikhao
AnswerSquare 10; $t_{n} = 7 \times 2^{n - 1}$ ; $t_{1} = 7$ , $t_{n} = 2 t_{n - 1}$ .
Daane ek GP banate hain jismein $a = 7$ aur $r = 2$ , isliye $t_{n} = 7 \times 2^{n - 1}$ .
$7 \times 2^{n - 1} = 3584$ se $2^{n - 1} = 512 = 2^{9}$ milta hai, isliye $n - 1 = 9$ aur $n = 10$ : square 10.
Recursive rule: $t_{1} = 7$ aur $n \geq 2$ ke liye $t_{n} = 2 t_{n - 1}$ .
5
Long answerMediumNCERT Exercise Set 8.3, Q5
Ek medhak seedhe raaste par kai hops lagata hai. Uska pehla hop 125 cm lamba hai, aur har hop pichhle hop ka 80% lamba hai.
1. Uska 5th hop kitna lamba hai?
  cm
  Answer dekho
  Answer 51.2 cm
  $125 \times {0.8}^{4} = 125 \times 0.4096 = 51.2$ cm.
2. 6 hops ke baad woh kitni door ja chuka hai?
  cm
  Answer dekho
  Answer 461.16 cm
  Hops 125, 100, 80, 64, 51.2 aur 40.96 cm hain, jinka sum 461.16 cm hai.
Short answerMediumNCERT Exercise Set 8.3, Q6
1 cm side wala ek square banao. Doosra square banao jiski side pehle ka diagonal ho, teesra jiski side doosre ka diagonal ho, aur aage isi tarah. Square ka diagonal uski side ka $\sqrt{2}$ guna hota hai, isliye sides $1, \sqrt{2}, 2, 2 \sqrt{2}, \dots$ cm hain. Kis square ki side 32 cm hai?
Answer dekho
Answer 11
Sides ek GP banati hain jismein $a = 1$ aur $r = \sqrt{2}$ , isliye $n$ th side $(\sqrt{2})^{n - 1}$ hai. Kyunki $32 = 2^{5} = (\sqrt{2})^{10}$ , hamein $n - 1 = 10$ chahiye: 11th square. (Unke areas 1, 2, 4, 8, … har baar double hote hain.)
7
Long answerHardNCERT Exercise Set 8.3, Q7
Fig. 8.18 ka cross fractal dekho, aur stage 0 par poore square ka area 1 square unit maano.
1. Stage 3 par shaded squares gino. Phir stage $n$ par ginti ke liye explicit rule aur recursive rule likho.
  Solution dikhao
  Answer125; $t_{n} = 5^{n}$ ; $t_{0} = 1$ , $t_{n} = 5 t_{n - 1}$ .
  Har shaded square ki jagah 5 chhote squares aa jaate hain, isliye har stage par ginti 5 se multiply hoti hai: 1, 5, 25, 125. Gintiyan 5 ki powers hain, isliye $t_{n} = 5^{n}$ , jahan $t_{0} = 1$ aur $t_{n} = 5 t_{n - 1}$ .
2. Kis stage par shaded squares ki ginti pehli baar 10 000 se upar jaati hai?
  Answer dekho
  Answer 6
  $5^{5} = 3125$ abhi bhi 10 000 se kam hai, lekin $5^{6} = 15 625$ usse upar hai: stage 6.
3. Stage 2 par shaded area kya hai? Stage $n$ par area ke liye explicit rule aur recursive rule likho.
  Solution dikhao
  Answer $\frac{25}{81}$ ; $s_{n} = {(\frac{5}{9})}^{n}$ ; $s_{0} = 1$ , $s_{n} = \frac{5}{9} s_{n - 1}$ .
  Har stage har shaded square ke 9 barabar hisson mein se 5 rakhta hai, isliye area $\frac{5}{9}$ se multiply hota hai: $1, \frac{5}{9}, \frac{25}{81}, \dots$ Isliye $s_{n} = {(\frac{5}{9})}^{n}$ , jahan $s_{0} = 1$ aur $s_{n} = \frac{5}{9} s_{n - 1}$ .
4. Kis stage par pehli baar square ka ek-dasvaan hissa se kam shaded bachta hai? Lambe samay mein shaded area ka kya hota hai?
  Solution dikhao
  AnswerStage 4; yeh 0 ki taraf ghatta jaata hai.
  ${(\frac{5}{9})}^{3} = \frac{125}{729} \approx 0.171$ 0.1 se zyada hai, lekin ${(\frac{5}{9})}^{4} = \frac{625}{6561} \approx 0.095$ kam hai: stage 4. Baad ka har stage area ko phir $\frac{5}{9}$ se multiply karta hai, isliye area 0 ke jitna chaho utna paas pahunch jaata hai, bhale hi squares ki ginti badhti rehti hai.
Quick answerMedium
AP 5, 10, 15, … aur GP 5, 10, 20, … ek hi tarah shuru hote hain. GP ka 10th term AP ke 10th term se kitna bada hai?
Answer dekho
Answer 2510
GP: $5 \times 2^{9} = 2560$ . AP: $5 + 9 \times 5 = 50$ . Difference $2560 - 50 = 2510$ hai.

Recap

Summary aur key points

Sequence ek tay order mein numbers ki list hai. Iske members terms hain, aur $t_{n}$ position $n$ wala term hai (Section 8.1).
Sequences finite ya infinite ho sakte hain, aur unke terms fractions, negative numbers ya koi bhi numbers ho sakte hain (Terms ko naam dena).
Triangular numbers running totals $1 + 2 + \dots + n$ hain; square numbers odd numbers ke running totals hain (Section 8.1).
Explicit rule $n$ se seedha $t_{n}$ deta hai. Koi number term hai ya nahi, yeh test karne ke liye $n$ ke liye solve karo aur check karo ki woh counting number hai (Section 8.2).
Recursive rule har term ko pehle ke terms se banata hai aur use starting terms chahiye, jaise Virahānka–Fibonacci rule $V_{n} = V_{n - 1} + V_{n - 2}$ mein (Section 8.3).
AP har step par same common difference $d$ jodta hai: $t_{n} = a + (n - 1) d$ , aur recursively $t_{n} = t_{n - 1} + d$ (Section 8.4).
AP ke points $(n, t_{n})$ ek straight line par hote hain jiska slope $d$ hai (Section 8.4.1).
$1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ , jo dono ends se numbers ke pairs banakar milta hai; yahi $n$ th triangular number bhi hai (Section 8.5).
Āryabhaṭa ka rule: AP ka sum uske pehle aur aakhri terms ka average guna terms ki ginti hai (Āryabhaṭa ka rule).
GP har step par same common ratio $r$ se multiply karta hai: $t_{n} = a r^{n - 1}$ , aur recursively $t_{n} = r t_{n - 1}$ (Section 8.6).
GP ke points line par nahi hote: $r > 1$ ke saath woh upar curve karte hain, aur $0 < r < 1$ ke saath 0 ki taraf girte hain (Section 8.6.2).
Sierpiński triangle jaise fractals GPs dete hain: stage $n$ par $3^{n}$ shaded triangles hote hain jinka total area ${(\frac{3}{4})}^{n}$ hai (Section 8.6.1).

Flash cards · 10

Key terms

1 / 10

Practice · 18 questions

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
Ek pahadi sadak barabar chadhti hai. 8 km stone par woh sea level se 1250 m upar hai, aur 20 km stone par sea level se 1850 m upar. Har kilometre par height utni hi badhti hai.
1. 35 km stone par sadak kitni oonchi hai?
  m
  Answer dekho
  Answer 2600 m
  $20 - 8 = 12$ km mein sadak $1850 - 1250 = 600$ m chadhti hai, isliye woh har km mein 50 m chadhti hai. 8 km stone se 35 km stone tak 27 km hai: $1250 + 27 \times 50 = 2600$ m.
2. Kis kilometre stone par sadak bilkul 2500 m oonchi hai?
  Answer dekho
  Answer 33
  $1250 + (n - 8) \times 50 = 2500$ se $n - 8 = 25$ milta hai, isliye yeh 33 km stone hai.
Short answerMediumNCERT End-of-chapter Exercises, Q2
Ek AP mein 5th term 2nd term ka teen guna hai, aur 8th term 45 hai. Pehle kuch terms likho.
Solution dikhao
Answer3, 9, 15, 21, …
$a + 4 d = 3 a + 3 d$ se $d = 2 a$ milta hai. Phir $a + 7 d = a + 14 a = 15 a = 45$ , isliye $a = 3$ aur $d = 6$ . AP 3, 9, 15, 21, 27, … hai. Check karo: $t_{2} = 9$ aur $t_{5} = 27 = 3 \times 9$ .
Short answerMediumNCERT End-of-chapter Exercises, Q3
Kitne three-digit numbers 9 se divide karne par remainder 2 chhodte hain?
Answer dekho
Answer 100
Sabse chhota $101 = 9 \times 11 + 2$ hai aur sabse bada $992 = 9 \times 110 + 2$ . $101 + (n - 1) \times 9 = 992$ solve karne par $n - 1 = 99$ milta hai, isliye aise 100 numbers hain.
Short answerMediumNCERT End-of-chapter Exercises, Q4
Ek seedhi sadak ke saath 0 m mark, 6 m mark, 12 m mark aur aage isi tarah lamps lage hain, har 6 m par ek. 20 m mark aur 290 m mark ke beech kitne lamps hain?
Answer dekho
Answer 45
Is hisse mein lamps 24, 30, …, 288 m par hain. $24 + (n - 1) \times 6 = 288$ solve karne par $n - 1 = 44$ milta hai, isliye 45 lamps hain.
Long answerHardNCERT End-of-chapter Exercises, Q5
Ek GP ka 3rd term 18 aur 5th term 162 hai. Har woh GP dhoondho jo fit hota hai.
Solution dikhao
Answer2, 6, 18, 54, 162, … ( $r = 3$ ke saath) ya 2, −6, 18, −54, 162, … ( $r = - 3$ ke saath).
$\frac{a r^{4}}{a r^{2}} = r^{2} = \frac{162}{18} = 9$ , isliye $r = 3$ ya $r = - 3$ .
Phir $a r^{2} = 18$ se $9 a = 18$ milta hai, isliye dono cases mein $a = 2$ .
Isliye GP 2, 6, 18, 54, 162, … ya 2, −6, 18, −54, 162, … hai. Dono fit hote hain, kyunki 3rd aur 5th terms mein $r$ ki even powers hain.
6
Long answerHardNCERT End-of-chapter Exercises, Q6
Kuch numbers ko do ya zyada consecutive natural numbers ke sum ki tarah likha ja sakta hai, jaise $9 = 4 + 5 = 2 + 3 + 4$ .
1. 45 ko har possible tareeke se aise sum ki tarah likho.
  Solution dikhao
  Answer $22 + 23$ , $14 + 15 + 16$ , $7 + 8 + 9 + 10 + 11$ , $5 + 6 + \dots + 10$ aur $1 + 2 + \dots + 9$ .
  Maan lo $k$ consecutive numbers $m$ se shuru hote hain. Āryabhaṭa ke rule se unka sum $k$ guna average $m + \frac{k - 1}{2}$ hai. Ise 45 ke barabar rakh kar double karne par $k (2 m + k - 1) = 90$ milta hai.
  Isliye $k$ aur $2 m + k - 1$ 90 ke do factors hain jinka product 90 hai. Unka difference, $2 m - 1$ , odd hai, isliye ek factor even hai aur doosra odd. Saath hi $2 m + k - 1 > k$ , kyunki $m \geq 1$ .
  $k \geq 2$ ke saath possible pairs $2 \times 45$ , $3 \times 30$ , $5 \times 18$ , $6 \times 15$ aur $9 \times 10$ hain. Inse $m = 22$ , 14, 7, 5 aur 1 milte hain: kul paanch tareeke.
2. Ab 32 try karo. Tumhe kya milta hai, aur kyun?
  Solution dikhao
  AnswerAisa nahi ho sakta.
  Ab $k (2 m + k - 1) = 64$ , aur do factors mein se ek odd hona chahiye. 64 ka akela odd factor 1 hai, lekin $k \geq 2$ aur $2 m + k - 1 > k$ , isliye koi bhi factor 1 nahi ho sakta. Isliye 32 do ya zyada consecutive natural numbers ka sum nahi hai. Yahi baat 2 ki har power ke liye sach hai.
7
Short answerMediumNCERT End-of-chapter Exercises, Q7
Ek naya video online share hota hai. Shuru mein 50 log use dekh chuke hain, aur dekhne walon ki ginti har ghante teen guna ho jaati hai.
1. 4 ghante ke baad kitne logon ne use dekha hai, aur $n$ ghante ke baad?
  Solution dikhao
  Answer4050 aur $50 \times 3^{n}$ .
  4 ghante ke baad: $50 \times 3^{4} = 50 \times 81 = 4050$ . $n$ ghante ke baad: $50 \times 3^{n}$ .
2. Kitne poore ghanton ke baad 1 lakh se zyada log use dekh chuke honge?
  hours
  Answer dekho
  Answer 7 hours
  $50 \times 3^{n} > 100 000$ ke liye $3^{n} > 2000$ chahiye. Kyunki $3^{6} = 729$ aur $3^{7} = 2187$ , yeh 7 ghante ke baad hota hai, jab $50 \times 2187 = 109 350$ log dekh chuke hote hain.
Long answerHardNCERT End-of-chapter Exercises, Q8
Ek badhte hue AP ke pehle teen terms ka sum 27 aur product 585 hai. Unhe nikaalo.
Solution dikhao
Answer5, 9, 13
Unka sum $(m - d) + m + (m + d) = 3 m = 27$ hai, isliye beech wala term $m = 9$ hai.
Unka product $(9 - d) \times 9 \times (9 + d) = 9 (81 - d^{2}) = 585$ hai, isliye $81 - d^{2} = 65$ aur $d^{2} = 16$ .
AP badh raha hai, isliye $d = 4$ aur terms 5, 9 aur 13 hain. Check karo: $5 + 9 + 13 = 27$ aur $5 \times 9 \times 13 = 585$ .
Short answerMediumNCERT End-of-chapter Exercises, Q9
Sabse chhota $n$ nikaalo jiske liye $1 + 2 + \dots + n$ 2026 se bada hai.
Answer dekho
Answer 64
$S_{63} = \frac{63 \times 64}{2} = 2016$ , jo kaafi nahi hai, lekin $S_{64} = \frac{64 \times 65}{2} = 2080 > 2026$ .
Short answerMediumNCERT End-of-chapter Exercises, Q10
Ek knockout chess tournament mein har round mein aadhe players bahar ho jaate hain. Round 1 mein 1024 players hain, round 2 mein 512, aur aage isi tarah. Kis round mein 8 players hain? Round $n$ mein players ki ginti ke liye explicit rule aur recursive rule likho.
Solution dikhao
AnswerRound 8; $t_{n} = 1024 \times {(\frac{1}{2})}^{n - 1}$ ; $t_{1} = 1024$ , $t_{n} = \frac{1}{2} t_{n - 1}$ .
Yahan $a = 1024$ aur $r = \frac{1}{2}$ , isliye $t_{n} = 1024 \times {(\frac{1}{2})}^{n - 1} = \frac{1024}{2^{n - 1}}$ .
$\frac{1024}{2^{n - 1}} = 8$ se $2^{n - 1} = 128 = 2^{7}$ milta hai, isliye $n = 8$ : round 8.
Recursive rule: $t_{1} = 1024$ aur $n \geq 2$ ke liye $t_{n} = \frac{1}{2} t_{n - 1}$ .
Long answerHardNCERT End-of-chapter Exercises, Q11
Ek cuboid box ki length, breadth aur height, isi order mein, ek GP banati hain. Box ka volume $1000 c m^{3}$ aur total surface area $700 c m^{2}$ hai. Uske teeno edges ki lengths nikaalo.
Solution dikhao
Answer5 cm, 10 cm aur 20 cm (kisi ek ya doosre order mein).
Volume: $\frac{a}{r} \times a \times a r = a^{3} = 1000$ , isliye $a = 10$ .
Surface area: $2 (\frac{a}{r} \times a + a \times a r + a r \times \frac{a}{r}) = 2 a^{2} (\frac{1}{r} + r + 1) = 200 (\frac{1}{r} + r + 1) = 700$ .
Isliye $\frac{1}{r} + r = 2.5$ . $2 r$ se multiply karne par $2 r^{2} - 5 r + 2 = 0$ milta hai, yaani $(2 r - 1) (r - 2) = 0$ , isliye $r = 2$ ya $r = \frac{1}{2}$ .
Dono tarah edges 5 cm, 10 cm aur 20 cm hain. Check karo: $5 \times 10 \times 20 = 1000$ aur $2 (50 + 200 + 100) = 700$ .
ProofHardNCERT End-of-chapter Exercises, Q12
Common ratio $r$ wala koi bhi GP $t_{1}, t_{2}, t_{3}, \dots$ lo. Prove karo ki padosi terms ke products, $t_{1} t_{2}, t_{2} t_{3}, t_{3} t_{4}, \dots$ , bhi ek GP banate hain, aur uska common ratio nikaalo.
Solution dikhao
AnswerWoh common ratio $r^{2}$ wala GP banate hain.
Maan lo GP ka first term $a$ hai, isliye $t_{n} = a r^{n - 1}$ aur $t_{n + 1} = a r^{n}$ . $n$ th product $p_{n} = t_{n} t_{n + 1} = a^{2} r^{2 n - 1}$ hai.
$\frac{p_{n + 1}}{p_{n}} = \frac{a^{2} r^{2 n + 1}}{a^{2} r^{2 n - 1}} = r^{2}$
Har $n$ ke liye ratio $r^{2}$ hai, isliye products common ratio $r^{2}$ wala GP banate hain. Jaise GP 2, 6, 18, 54 se products 12, 108 aur 972 milte hain, har ek pichhle ka 9 guna.
Long answerMediumNCERT End-of-chapter Exercises, Q13
Ek GP mein 1st aur 3rd terms ka sum 50 hai, aur 2nd aur 4th terms ka sum 150. GP nikaalo.
Solution dikhao
Answer5, 15, 45, 135, …
$a + a r^{2} = a (1 + r^{2}) = 50$ aur $a r + a r^{3} = a r (1 + r^{2}) = 150$ .
Doosri equation ko pehli se divide karne par $r = 3$ milta hai (factor $1 + r^{2}$ kabhi 0 nahi hota).
Phir $a (1 + 9) = 50$ , isliye $a = 5$ . GP 5, 15, 45, 135, … hai. Check karo: $5 + 45 = 50$ aur $15 + 135 = 150$ .
14
Long answerHardNCERT End-of-chapter Exercises, Q14
Maan lo $R_{1} = 2$ aur $n \geq 2$ ke liye $R_{n} = 2 (R_{1} + R_{2} + \dots + R_{n - 1})$ , yaani ab tak ke saare terms ke total ka double.
1. $R_{1}, R_{2}, \dots, R_{8}$ nikaalo.
  Solution dikhao
  Answer2, 4, 12, 36, 108, 324, 972, 2916
  $R_{2} = 2 \times 2 = 4$ , $R_{3} = 2 (2 + 4) = 12$ , $R_{4} = 2 (2 + 4 + 12) = 36$ , $R_{5} = 2 (54) = 108$ , $R_{6} = 2 (162) = 324$ , $R_{7} = 2 (486) = 972$ aur $R_{8} = 2 (1458) = 2916$ .
2. Ek simple recursive rule aur ek explicit rule dhoondho. Woh kis term se aage kaam karte hain?
  Solution dikhao
  Answer $n \geq 3$ ke liye $R_{n} = 3 R_{n - 1}$ ; $n \geq 2$ ke liye $R_{n} = 4 \times 3^{n - 2}$ (aur $R_{1} = 2$ alag se).
  $n \geq 3$ ke liye, $R_{n}$ ke rule ke andar ka total $R_{n - 1}$ wala total plus ek aur term, $R_{n - 1}$ , hai. Purane total ka double khud $R_{n - 1}$ hai, isliye $R_{n} = R_{n - 1} + 2 R_{n - 1} = 3 R_{n - 1}$ .
  Yeh shuruaat mein fail hota hai: $R_{2} = 4$ , lekin $3 R_{1} = 6$ . Isliye $R_{2}$ se aage ke terms first term 4 aur ratio 3 wala GP banate hain, jisse $n \geq 2$ ke liye $R_{n} = 4 \times 3^{n - 2}$ milta hai, jabki $R_{1} = 2$ alag rehta hai. Check karo: $R_{8} = 4 \times 3^{6} = 2916$ .
15
Long answerHardNCERT End-of-chapter Exercises, Q15
Maan lo $V_{1}, V_{2}, V_{3}, \dots$ Virahānka–Fibonacci sequence 1, 2, 3, 5, 8, … hai, aur $Q_{n} = V_{1} + V_{2} + \dots + V_{n}$ uske running totals hain.
1. $Q_{1}$ se $Q_{6}$ tak nikaalo aur unhe Virahānka–Fibonacci sequence ke terms se compare karo. Tumne kya notice kiya?
  Solution dikhao
  Answer1, 3, 6, 11, 19, 32; har ek do jagah aage wale term se 2 kam hai: $Q_{n} = V_{n + 2} - 2$ .
  $Q_{1} = 1$ , $Q_{2} = 3$ , $Q_{3} = 6$ , $Q_{4} = 11$ , $Q_{5} = 19$ aur $Q_{6} = 32$ . $V_{3}$ se aage Virahānka–Fibonacci terms 3, 5, 8, 13, 21 aur 34 hain, isliye $Q_{1} = 3 - 2$ , $Q_{2} = 5 - 2$ , $Q_{3} = 8 - 2$ , $Q_{4} = 13 - 2$ , $Q_{5} = 21 - 2$ aur $Q_{6} = 34 - 2$ .
2. Samjhao ki $Q_{n} = V_{n + 2} - 2$ har $n$ ke liye kyun kaam karta rehta hai.
  Solution dikhao
  AnswerHar naya total ek aur term jodta hai, aur baaki kaam Virahānka–Fibonacci rule kar deta hai.
  Yeh $n = 1$ ke liye kaam karta hai: $Q_{1} = 1 = V_{3} - 2$ . Ab maan lo yeh kisi $n$ ke liye kaam karta hai. Tab rule $V_{n + 3} = V_{n + 2} + V_{n + 1}$ use karke $Q_{n + 1} = Q_{n} + V_{n + 1} = V_{n + 2} - 2 + V_{n + 1} = V_{n + 3} - 2$ . Isliye jab bhi yeh ek $n$ ke liye kaam karta hai, agle ke liye bhi karta hai; $n = 1$ se shuru karke yeh har $n$ ke liye kaam karta hai.
Quick answerEasy
Kaun sa sequence AP aur GP dono hai?
1, 2, 4, 8, …2, 4, 6, 8, …1, −1, 1, −1, …4, 4, 4, 4, …
Answer dekho
Answer (D) 4, 4, 4, 4, …
Constant sequence ka common difference 0 aur common ratio 1 hota hai. Sequence 1, −1, 1, −1 $r = - 1$ wala GP hai, lekin uske differences $- 2, 2, - 2$ constant nahi hain.
Quick answerEasy
Ek AP mein $t_{1} = 7$ aur $t_{2} = 11$ hai, isliye $t_{100} =$ .
Answer dekho
Answer 403
$d = 4$ , isliye $t_{100} = 7 + 99 \times 4 = 403$ .
18
Quick answerMedium
In numbers ko badhte order mein rakho.
1. $2^{10}$
2. $1 + 2 + \dots + 10$
3. AP 5, 10, 15, … ka 10th term
4. GP 1, 2, 4, … ka 10th term
Answer dekho
Answer
AP 5, 10, 15, … ka 10th term
$1 + 2 + \dots + 10$
GP 1, 2, 4, … ka 10th term
$2^{10}$
Yeh 50, 55, $2^{9} = 512$ aur 1024 hain.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise Set 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise Set 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise Set 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise Set 8.1, Q4	Practice 8.1 · Q4
NCERT Exercise Set 8.1, Q5	Practice 8.1 · Q5
NCERT Exercise Set 8.1, Q6	Practice 8.1 · Q6
NCERT Exercise Set 8.2, Q1	Practice 8.2 · Q1
NCERT Exercise Set 8.2, Q2	Practice 8.2 · Q2
NCERT Exercise Set 8.2, Q3	Practice 8.2 · Q3
NCERT Exercise Set 8.2, Q4	Practice 8.2 · Q4
NCERT Exercise Set 8.2, Q5	Practice 8.2 · Q5
NCERT Exercise Set 8.2, Q6	Practice 8.2 · Q6
NCERT Exercise Set 8.2, Q7	Practice 8.2 · Q7
NCERT Exercise Set 8.3, Q1	Practice 8.3 · Q1
NCERT Exercise Set 8.3, Q2	Practice 8.3 · Q2
NCERT Exercise Set 8.3, Q3	Practice 8.3 · Q3
NCERT Exercise Set 8.3, Q4	Practice 8.3 · Q4
NCERT Exercise Set 8.3, Q5	Practice 8.3 · Q5
NCERT Exercise Set 8.3, Q6	Practice 8.3 · Q6
NCERT Exercise Set 8.3, Q7	Practice 8.3 · Q7
NCERT End-of-chapter Exercises, Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises, Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises, Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises, Q4	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises, Q5	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises, Q6	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises, Q7	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises, Q8	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises, Q9	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises, Q10	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises, Q11	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises, Q12	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises, Q13	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises, Q14	Chapter ke end ki exercises · Q14
NCERT End-of-chapter Exercises, Q15	Chapter ke end ki exercises · Q15

Quick check: MCQs

Terms ko naam dena4 questions
8.2 Sequence ka Explicit Rule4 questions
Virahānka–Fibonacci sequence4 questions
8.4 Arithmetic Progressions4 questions
8.4.1 AP ko Visualise karna3 questions
8.5 Pehle n Natural Numbers ka Sum4 questions
8.6 Geometric Progressions4 questions
8.6.2 GP ko Visualise karna3 questions

Glossary · 12

Key terms (English + Hindi)

arithmetic progressionसमांतर श्रेढ़ीsamantar shredhi: Aisa sequence jo har step par same number $d$ jodta hai: $a, a + d, a + 2 d, \dots$

common differenceसार्व अंतरsarva antar: AP ke har step par joda jaane wala fixed number $d$ ; $d = t_{2} - t_{1}$ .Chapter mein pehli baar yahan aaya →
common ratioसार्व अनुपातsarva anupat: GP ka fixed multiplier $r$ ; $r = t_{2} \div t_{1}$ .Chapter mein pehli baar yahan aaya →

explicit rule: Aisa formula jo $n$ se seedha $n$ th term deta hai, jaise $t_{n} = 7 n + 3$ .Chapter mein pehli baar yahan aaya →

finite sequenceपरिमित अनुक्रमparimit anukram: Aisa sequence jo kuch terms ke baad ruk jaata hai. Jo kabhi nahi rukta, woh infinite sequence hai.Chapter mein pehli baar yahan aaya →
fractal: Aisi shape jo apni hi chhoti aur chhoti copies se bani ho, jaise Sierpiński triangle.Chapter mein pehli baar yahan aaya →

geometric progressionगुणोत्तर श्रेढ़ीgunottar shredhi: Aisa sequence jo har step par same number $r$ se multiply karta hai: $a, a r, a r^{2}, \dots$ Chapter mein pehli baar yahan aaya →

recursive ruleपुनरावर्ती नियमpunaravarti niyam: Aisa rule jo har term pehle ke terms se deta hai, saath mein starting term(s) bhi, jaise $t_{1} = 6$ , $t_{n} = t_{n - 1} + 5$ .Chapter mein pehli baar yahan aaya →

sequenceअनुक्रमanukram: Ek tay order mein numbers ki list, jaise 3, 10, 17, 24, 31.Chapter mein pehli baar yahan aaya →

termपदpad: Sequence ka ek number. Position $n$ wala term $t_{n}$ likha jaata hai.Chapter mein pehli baar yahan aaya →
triangular numberत्रिभुजीय संख्याtribhujiya sankhya: $1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ form ka number: 1, 3, 6, 10, 15, …Chapter mein pehli baar yahan aaya →

Virahānka–Fibonacci sequence: Har term apne se pehle ke do terms ka sum hai, 1, 2 se shuru: yaani 1, 2, 3, 5, 8, 13, …Chapter mein pehli baar yahan aaya →

8.1 Sequences ka Introduction

Terms ko naam dena

8.2 Sequence ka Explicit Rule

8.3 Sequence ka Recursive Rule

Virahānka–Fibonacci sequence

8.4 Arithmetic Progressions

8.4.1 AP ko Visualise karna

8.5 Pehle n Natural Numbers ka Sum

8.6 Geometric Progressions

8.6.1 Fractals ke saath Maza

8.6.2 GP ko Visualise karna

Apna NCERT exercise question dhoondo

Quick check: MCQs

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