Sequences and Series: Class 11 Maths Chapter 8 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

80 min
1 simulation
61 practice questions
20 MCQs

Is chapter mein aap kya seekhenge

Sequence ko natural numbers par ek function ki tarah dekhna, nth-term formula ya recurrence relation se uske terms nikaalna, aur finite aur infinite sequences mein fark pehchaanna.
Series ko sigma notation mein likhna aur series ko uske sum se alag pehchaanna.
Geometric progression ko pehchaanna, uska common ratio nikaalna aur general term a_n = ar^(n−1) use karna.
G.P. ke pehle n terms ka sum S_n derive karna aur use sums, terms ki sankhya aur di gayi conditions se terms nikaalne mein use karna.
Do positive numbers ka geometric mean nikaalna aur unke beech kitne bhi geometric means insert karna.
Do positive numbers ke liye A.M. ≥ G.M. prove karna aur ise means se numbers nikaalne aur sabse bada product nikaalne mein use karna.
Growth, decay, compound interest, depreciation aur instalment wale problems ko geometric aur arithmetic progressions se model karna.

NCERT Mathematics Class 11, Chapter 8, pp. 135–150 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

8.1 Introduction

Ek batter har over mein jo runs banata hai, garmi ke ek din Nagpur ka har ghante ka temperature, ya novel padhte hue har raat tum jis page tak pahunchte ho, inhe note karo. Har baar tumhe numbers ek pakke order mein milte hain: pehla, doosra, teesra, aur aage bhi. Aisi ordered lists sequences kehlati hain, aur yahi is chapter ka topic hai.

Is chapter mein do rules sabse zaroori hain. Har baar same number jodne se arithmetic progression (A.P.) banti hai, jo Class 10 se jaani-pehchaani hai. Har baar same number se multiply karne se geometric progression (G.P.) banti hai, jo hamara main topic hai. Aise rule-based sequences ko progressions kehte hain. G.P. ke liye hum koi bhi term aur bahut saare terms ka sum nikaalenge, geometric mean se milenge, aur dekhenge ki do positive numbers ka arithmetic mean unke geometric mean se kabhi chhota kyun nahi ho sakta.

8.2 Sequences

Kaagaz ki ek sheet ko aadha fold karo, to usmein 2 layers ho jaati hain. Phir se fold karo: 4 layers. Phir 8, phir 16. Saat folds se 128 layers banti hain, lagbhag ek notebook jitna mota stack. Counts 2, 4, 8, 16, 32, 64, 128 ek fixed order mein aate hain, har fold ke liye ek, aur yahi order inhe sequence banata hai.

Ab square numbers ki list banao: 1, 4, 9, 16, 25, …. Yeh list kabhi khatam nahi hoti, kyunki har square ke baad ek aur bada square hota hai.

Sequence ka har number ek term hai. Hum terms ko unki positions se naam dete hain: $a_{1}$ pehla term hai, $a_{2}$ doosra, aur general mein $a_{n}$ position $n$ wala term hai. Subscript batata hai ki term kahan khada hai; $a_{n}$ ki value batati hai ki wahan kya khada hai. Folding sequence mein $a_{1} = 2$ , $a_{3} = 8$ aur $a_{7} = 128$ . Agar $n$ ke terms mein $a_{n}$ ka koi formula ho, to woh sequence ka general term (ya $n$ th term) hai.

Jis sequence mein terms ki sankhya fixed ho, woh finite hai, jaise saat layer counts ya March ke Mondays ki dates. Jo sequence bina ruke chalta rahe, woh infinite hai, jaise square numbers ya 7 ke multiples.

Aksar ek hi formula har term deta hai. 5 ke multiples 5, 10, 15, 20, … hain, aur position $n$ wala term $5 n$ hai, isliye $a_{n} = 5 n$ . Sequence 3, 7, 11, 15, … 3 se shuru hota hai aur har baar 4 jodta hai, isliye $a_{n} = 4 n - 1$ ; check karo ki $a_{1} = 3$ aur $a_{4} = 15$ . Fig. 8.1 is rule ko positions aur values ki matching ke roop mein dikhata hai.

3/3Graph alag-alag dots $(n, a_{n})$ ka set hai: $n = 2.5$ par koi term nahi hai.

Fig. 8.1Sequence har position

n

ko ek term

a_{n}

se jodta hai.

Left mein paanch boxes ka ek column hai jismein positions 1, 2, 3, 4 aur 5 hain. Pehle step mein sirf yeh positions dikhti hain. Doosre step mein har position se ek arrow boxes ke doosre column tak jaata hai jismein terms 3, 7, 11, 15 aur 19 hain, aur right mein rule a sub n equals 4n minus 1 dikhta hai. Teesre step mein right mein ek graph aata hai jiske horizontal axis par n 1 se 5 tak hai aur vertical axis par term ki value 20 tak; paanchon pairs alag dots ki tarah barabar steps mein upar chadhte hain, aur unke beech kuch nahi hai.

Kabhi-kabhi yeh batana aasaan hota hai ki har term apne pichhle terms se kaise banta hai. Aisa rule, shuru ke terms ke saath, ek recurrence relation hai. Sabse mashhoor example Fibonacci sequence hai: $a_{1} = a_{2} = 1$ se shuru karo aur $n \geq 3$ ke liye yeh use karo

Fibonacci rule

a_{n} = a_{n - 1} + a_{n - 2}

Isse 1, 1, 2, 3, 5, 8, 13, 21, … milta hai, har term apne pichhle do terms ka sum hai. Fig. 8.2 mein har naya term ek square ki side banta hai jo ab tak bane rectangle ki lambi edge ke saath bilkul fit hota hai.

4/4Side $3 + 5 = 8$ ka square ek $13 \times 8$ rectangle poora karta hai.

Fig. 8.2Har nayi side pichhli do sides ka sum hai, isliye har square rectangle ki lambi edge par fit hota hai.

Pehle step mein side 1 ke do chhote squares agal-bagal hain, aur neeche ki line mein 1, 1 likha hai. Doosre step mein unke neeche side 2 ka square judta hai, jo unki milkar chaudai ke barabar hai, aur line mein 1, 1, 2 likha hai. Teesre step mein pehle teen ke bagal mein side 3 ka square judta hai, phir sabke upar side 5 ka square, aur line mein 1, 1, 2, 3, 5 likha hai. Chauthe step mein side 8 ka bada square left side bharta hai, 13 by 8 ka rectangle poora hota hai, aur line mein 1, 1, 2, 3, 5, 8 likha hai.

Har sequence ke saath formula nahi aata. $π$ ke digits, order mein padhne par, 3, 1, 4, 1, 5, 9, 2, 6, … dete hain, ek infinite sequence jiske $n$ th digit ka koi saaf formula nahi hai; hum rule sirf shabdon mein bata sakte hain. Primes 2, 3, 5, 7, 11, … bhi aise hi hain. Har sequence ko bas itna chahiye ki har position par kaun sa number hai, yeh tay karne ka ek pakka tareeka ho.

Definition

Sequence

Ek function jo poore $𝐍$ par (infinite sequence) ya natural numbers ke shuru wale block ${1, 2, 3, \dots, k}$ par (finite sequence jismein $k$ terms hain) defined ho. Position $n$ par iski value $a_{n}$ likhi jaati hai, ya function notation mein $a (n)$ .

Example $a (n) = n^{2}$ se infinite sequence $1, 4, 9, 16, \dots$ milta hai

Sequence ko function ki tarah dekhne se yeh chapter Chapter 2 se judta hai: input ek position hai aur output ek term. Isiliye Fig. 8.1 ka graph alag-alag dots ka set hai, har natural number ke upar ek, na ki ek continuous line.

8.3 Series

Saal ke pehle week mein ₹10 bachao, doosre mein ₹20, teesre mein ₹30, aur aise hi aage. Weekly amounts ek sequence banate hain, lekin sawaal '8 weeks ke baad kitna?' $10 + 20 + \dots + 80$ ke baare mein hai. Sequence ke terms plus signs se jude hon to series banti hai; usmein utne hi terms hote hain jitne us sequence mein jisse woh bani hai, isliye woh finite ya infinite ho sakti hai.

Lambe sums likhna thakaane wala hai, isliye hum sigma notation use karte hain. Capital Greek letter $Σ$ (sigma, Greek S, 'sum' ke liye) poore sum ko ek symbol mein pack kar deta hai:

sigma notation

\begin{aligned} \sum_{k = 1}^{n} a_{k} \\ = a_{1} + a_{2} + \dots + a_{n} \end{aligned}

(8.1)

jahan

$k$: index, jo $1, 2, \dots, n$ se hokar chalta hai
$a_{k}$: position $k$ wala term

Ise aise padho: ' $k$ ke 1 se $n$ tak ke liye $a_{k}$ ka sum'. Letter $k$ sirf ek counter hai, isliye $\sum_{j = 1}^{n} a_{j}$ ka matlab bhi wahi sum hai. Jaise, $\sum_{k = 1}^{5} (2 k - 1)$ ka matlab $1 + 3 + 5 + 7 + 9$ hai, aur upar wali savings $\sum_{k = 1}^{8} 10 k$ hain. Fig. 8.3 dikhata hai ki pehle paanch odd numbers ka sum $5^{2} = 25$ kyun hai.

5/59 ki layer se $5 \times 5$ square banta hai: $\sum_{k = 1}^{5} (2 k - 1) = 25.$

Fig. 8.3Har odd number ek L-shaped layer hai, isliye pehle

n

odd numbers ek

n \times n

square bhar dete hain.

Left mein ek akela unit square kone mein hai aur right mein line 1 = 1 likhi hai. Doosre step mein 3 squares ki L-shaped layer uske chaaron or lipatti hai, 2 by 2 square banta hai, aur 1 + 3 = 4 dikhta hai. Teesre step mein 5 ki layer se 3 by 3 square banta hai aur 1 + 3 + 5 = 9 dikhta hai. Chauthe step mein 7 ki layer se 4 by 4 square banta hai aur 1 + 3 + 5 + 7 = 16 dikhta hai. Paanchve step mein 9 ki layer 5 by 5 square poora karti hai, 1 + 3 + 5 + 7 + 9 = 25 dikhta hai, aur neeche sum ka sigma form likha hota hai.

Do ideas ko alag rakho. Series ek expression hai: plus signs se jude terms, jaise $10 + 20 + 30 + 40$ . Uska sum woh number hai jo unhe sach mein jodne par milta hai, yahan 100. 'Series ka sum nikaalo' hamesha wahi number maangta hai.

Example 8.1NCERT Example 1 · p. 137

Formula se terms

Har sequence ke pehle teen terms likho: (i) $a_{n} = 5 - 3 n$ ; (ii) $a_{n} = \frac{n^{2}}{n + 1}$ .

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

Example 8.2NCERT Example 2 · p. 138

List mein bahut aage ka term

Is sequence ka 25th term nikaalo: $a_{n} = \frac{(- 1)^{n} (n^{2} - 1)}{4 n} .$

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

Example 8.3NCERT Example 3 · p. 138

Recurrence se terms

Ek sequence mein $a_{1} = 64$ aur $n \geq 2$ ke liye $a_{n} = - \frac{1}{2} a_{n - 1}$ hai. Iske pehle paanch terms aur unki series likho.

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

Quick check

3 mein se 1

\sum_{k = 1}^{4} k^{2}

evaluate karo.

Answer dekho

Answer 30

1 + 4 + 9 + 16 = 30

. Sum

1 + 2 + 3 + 4 = 10

ka square karke 100 lena aam galti hai: square har term par lagta hai.

Practice · 8 questions

Practice 8.1: answers ke saath

1
Short answerEasyNCERT Exercise 8.1, Q1–Q3
Har sequence ke pehle paanch terms likho.
1. $a_{n} = n^{2} - 2 n$
  Solution dikhao
  $- 1, 0, 3, 8, 15$
2. $a_{n} = \frac{2 n}{n + 2}$
  Solution dikhao
  $\frac{2}{3}, 1, \frac{6}{5}, \frac{4}{3}, \frac{10}{7}$
3. $a_{n} = 3^{n} - 1$
  Solution dikhao
  $2, 8, 26, 80, 242$
2
Short answerEasyNCERT Exercise 8.1, Q4–Q6
Har sequence ke pehle paanch terms likho.
1. $a_{n} = \frac{3 n - 2}{4}$
  Solution dikhao
  $\frac{1}{4}, 1, \frac{7}{4}, \frac{5}{2}, \frac{13}{4}$
2. $a_{n} = (- 1)^{n} 2^{n + 1}$
  Solution dikhao
  $- 4, 8, - 16, 32, - 64$
3. $a_{n} = \frac{n (n^{2} - 1)}{6}$
  Solution dikhao
  $0, 1, 4, 10, 20$ . Koi term 0 bhi ho sakta hai.
3
Quick answerEasyNCERT Exercise 8.1, Q7
Sequence $a_{n} = 7 n - 5$ ke liye neeche diye terms nikaalo.
1. $a_{12}$
  Answer dekho
  Answer 79
  $7 (12) - 5 = 79$ .
2. $a_{30}$
  Answer dekho
  Answer 205
  $7 (30) - 5 = 205$ .
Quick answerEasyNCERT Exercise 8.1, Q8
$a_{n} = \frac{2^{n}}{n^{2}}$ ke liye $a_{8}$ nikaalo.
Answer dekho
Answer 4
$a_{8} = \frac{256}{64} = 4$ .
Quick answerEasyNCERT Exercise 8.1, Q9
$a_{n} = (- 1)^{n} n^{2}$ ke liye $a_{11}$ nikaalo.
Answer dekho
Answer -121
11 odd hai, isliye $(- 1)^{11} = - 1$ aur $a_{11} = - 121$ .
Quick answerEasyNCERT Exercise 8.1, Q10
$a_{n} = \frac{n^{2} - 4}{n + 5},$ ke liye $a_{15}$ ko decimal mein nikaalo.
Answer dekho
Answer 11.05
$a_{15} = \frac{221}{20} = 11.05.$
7
Short answerMediumNCERT Exercise 8.1, Q11–Q13
Har sequence ke pehle paanch terms aur unki series likho.
1. $a_{1} = 2$ , $n \geq 2$ ke liye $a_{n} = 2 a_{n - 1} + 3$
  Solution dikhao
  $a_{2} = 7$ , $a_{3} = 17$ , $a_{4} = 37$ , $a_{5} = 77$ . Series: $2 + 7 + 17 + 37 + 77 + \dots$
2. $a_{1} = 3$ , $n \geq 2$ ke liye $a_{n} = \frac{(n - 1) a_{n - 1}}{n}$
  Solution dikhao
  $a_{2} = \frac{3}{2}$ , $a_{3} = 1$ , $a_{4} = \frac{3}{4}$ , $a_{5} = \frac{3}{5}$ ; asal mein $a_{n} = \frac{3}{n}$ . Series: $3 + \frac{3}{2} + 1 + \frac{3}{4} + \frac{3}{5} + \dots$
3. $a_{1} = 1$ , $a_{2} = 4$ , $n \geq 3$ ke liye $a_{n} = a_{n - 1} - a_{n - 2}$
  Solution dikhao
  $a_{3} = 3$ , $a_{4} = - 1$ , $a_{5} = - 4$ . Series: $1 + 4 + 3 - 1 - 4 + \dots .$ (Aage badhte raho: terms har 6 jagah ke baad repeat hote hain.)
Short answerMediumNCERT Exercise 8.1, Q14
Sequence $b_{1} = 1$ , $b_{2} = 3$ , $b_{n} = b_{n - 1} + b_{n - 2}$ ( $n \geq 3$ ) Fibonacci rule hi use karta hai, bas starting values alag hain. $n = 1, 2, 3, 4, 5$ ke liye $\frac{b_{n + 1}}{b_{n}}$ nikaalo.
Solution dikhao
Terms 1, 3, 4, 7, 11, 18 hain.
Ratios $3$ , $\frac{4}{3}$ , $\frac{7}{4}$ , $\frac{11}{7}$ aur $\frac{18}{11} \approx 1.636.$ hain. Yeh ek limit ke upar-neeche jhoolte hue $1.618$ ke paas aate jaate hain, jo Fibonacci numbers wala hi golden ratio hai.

8.4 Geometric Progression (G.P.)

Badhne ke do tareeke compare karo. 2 se shuru karo aur 3 jodte raho: 2, 5, 8, 11, 14, 17. 1 se shuru karo aur 2 se multiply karte raho: 1, 2, 4, 8, 16, 32. Pehli list aage se shuru hoti hai, lekin chhathe term tak doosri lagbhag do guna badi ho jaati hai (Fig. 8.4). Baar-baar multiply karna hi woh tareeka hai jisse compound interest par paisa badhta hai, forward kiya gaya message failta hai, aur blood mein kisi medicine ka level girta hai.

3/3 $n = 6$ tak doubling wali list (32) adding wali list (17) se bahut aage hai.

Fig. 8.4Fixed amount jodne se barabar steps milte hain; fixed factor se multiply karne se steps badhte hi jaate hain.

n = 1 se 6 ke upar bars ke chhah pairs khade hain. Pehle step mein sirf 3 jodne wali list ki solid blue bars dikhti hain, jo 2 se 17 tak barabar steps mein upar jaati hain. Doosre step mein unke bagal mein 2 se multiply karne wali list ki hatched orange bars aati hain: 1, 2, 4, 8, 16, 32, pehle chhoti aur phir tezi se upar. Teesre step mein ek dashed ring n = 6 ko mark karti hai, jahan 32, 17 se bahut upar hai.

Do aur examples. 8 m se giraayi gayi ball har baar apni pichhli height ke aadhe tak uchhalti hai, aur 4 m, 2 m, 1 m, 0.5 m, … tak pahunchti hai. Ek message jise har insaan teen naye logon ko forward karta hai, lagaatar stages mein 1, 3, 9, 27, … logon tak pahunchta hai. Har list mein kisi bhi term ko uske pichhle term se divide karne par hamesha same number milta hai: doubling list ke liye 2, ball ke liye $\frac{1}{2}$ aur message ke liye 3.

Definition

Geometric progression (G.P.)

Ek sequence geometric progression hai jab uska koi bhi term 0 na ho aur har term ko uske pichhle term se divide karne par ek hi number $r$ mile, yaani sabhi $k \geq 1$ ke liye $\frac{a_{k + 1}}{a_{k}} = r$ . Is number ko common ratio kehte hain.

Example $5$ , $- 15$ , $45$ , $- 135$ , … mein har term pichhle ka $- 3$ guna hai, isliye $r = - 3$ .

Agar first term $a$ hai, to G.P. $a, a r, a r^{2}, a r^{3}, \dots$ hoti hai. Terms non-zero hone chahiye kyunki hum unse divide karte hain; isi wajah se $r = 0$ allowed nahi hai, kyunki isse aage ka har term 0 ho jaayega. Ratio negative ho sakta hai, jisse signs baari-baari badalte hain, ya fraction ho sakta hai, jisse terms chhote hote jaate hain.

first term: $a$
common ratio: $r$
terms ki sankhya: $n$
last term: $l$ (ya $a_{n}$ )
pehle $n$ terms ka sum: $S_{n}$

Try karke dekho

Fold karo aur gino

Tumhe chahiye

newspaper ki ek sheet
ek ruler

Pehle guess karo

Agar ek sheet ko 10 baar aadha fold kiya ja sake, to usmein kitni layers hongi?

201005121024

Sheet ko aadha fold karo aur layers gino. Jitni baar fold kar sako, utni baar repeat karo.
Har count table mein likho, phir har count ko uske pichhle count se divide karo.
Aakhri fold ke baad stack ki thickness naapo.

Table 8.1Folds aur layers

Folds	Layers
1
2
3
4
5

Tumhari entries apne aap save hoti hain.

Pairs $(n, a_{n})$ ko plot karne se fark ek nazar mein dikh jaata hai. A.P. ke terms barabar steps se badalte hain, isliye unke dots ek straight line par hote hain. G.P. ke terms barabar factors se badalte hain, isliye unke dots kisi bhi line se door mudte hain. Neeche try karo.

SimulationA.P. ya G.P.?

Dono axes par 0 se 9 tak ka grid, jismein chaar draggable points a1 se a4 (1, 1), (2, 2), (3, 3) aur (4, 4) par hain, har ek apne coordinates dikhata hai aur whole numbers par snap hota hai. Shuru mein yeh A.P. 1, 2, 3, 4 hain, jo ek straight line par hain. Points ko vertically drag karne se terms badalte hain, isliye learner 1, 2, 4, 8 ya 8, 4, 2, 1 jaisi G.P. bana sakta hai aur dekh sakta hai ki dots line mein aane ki jagah mudte hain.

Fig. 8.5Har dot ek pair

(n, a_{n})

hai. Har dot ko uske apne

n

ke upar rakho aur term badalne ke liye use upar ya neeche drag karo.

Try karo

$a_{3}$ ko 4 tak aur $a_{4}$ ko 8 tak upar drag karke G.P. $1, 2, 4, 8$ banao. Kya chaaron dots ab bhi ek straight line par hain?
Ab ratio $\frac{1}{2}$ wali G.P. $8, 4, 2, 1$ banao. Dots ki chain kis taraf mudti hai?
Kya positive terms wali G.P. ke teen dots kabhi ek line par ho sakte hain? Heights 3, 3, 3 try karo. Yeh kaun sa ratio hai?

8.4.1 G.P. ka general term

Maan lo ek G.P. ka first term $a$ aur common ratio $r$ hai. Second term $a$ ko $r$ se ek baar multiply karke milta hai, third $a$ ko $r$ se do baar, aur fourth teen baar (Fig. 8.6). Position $n$ tak pahunchne ke liye hum position 1 se shuru karke $n - 1$ jumps lete hain, har jump $r$ se ek multiplication hai. Isse (8.2) milta hai:

3/3Position $n$ tak pahunchne mein $n - 1$ jumps lagte hain, isliye $a_{n} = a r^{n - 1}$ .

Fig. 8.6

r

ki power jumps ginti hai, jo position se ek kam hai.

Pehle step mein 1st label wala ek akela box first term a rakhta hai. Doosre step mein uske right mein aur boxes aate hain, jin par 2nd, 3rd, 4th aur nth label hai, aur jinmein ar, ar squared, ar cubed aur, dots wale gap ke baad, ar to the power n minus 1 hai; pehle boxes ke neeche ke curved arrows par times r label hai. Teesre step mein poori row ke neeche ek bracket par n minus 1 jumps, har baar times r, likha hai, jo dikhata hai ki r ki power position se ek kam kyun hai.

G.P. ka nth term

a_{n} = a r^{n - 1}

(8.2)

Teen baatein. G.P. ruk bhi sakti hai ya hamesha chalti bhi reh sakti hai; $n$ terms wali finite G.P. $a r^{n - 1}$ par khatam hoti hai. G.P. ke terms jodne se geometric series banti hai, jo usi hisaab se finite ya infinite hoti hai. Aur kisi bhi answer ka quick check: $r$ ki power hamesha position se ek kam hoti hai.

8.4.2 G.P. ke n terms ka sum

Terms ko ek-ek karke jodna chhah terms ke liye theek hai, lekin saath terms ke liye bekaar. Ek smart trick kisi bhi $n$ ke liye kaam karti hai. First term $a$ aur ratio $r$ wali G.P. ke pehle $n$ terms ke sum ko $S_{n}$ likho.

DerivationPehle n terms ka sum

Case $r = 1$ . Har term $a$ ke barabar hai, isliye $S_{n} = a + a + \dots + a = n a .$

Case $r \neq 1$ . Sum likho, phir har term ko $r$ se multiply karo:

\begin{aligned} S_{n} = a + a r + \dots \\ + a r^{n - 1} \end{aligned}

\begin{aligned} r S_{n} = a r + \dots \\ + a r^{n - 1} + a r^{n} \end{aligned}

Dono lines mein $a r$ se $a r^{n - 1}$ tak ka har term common hai (Fig. 8.7). Pehli line se doosri ghatane par yeh sab cancel ho jaate hain; sirf upar ka first term aur neeche ka last term bachte hain:

\begin{aligned} (1 - r) S_{n} \\ = a - a r^{n} \end{aligned}

Kyunki $r \neq 1$ , hum $1 - r$ se divide kar sakte hain:

n terms ka sum, r ≠ 1

\begin{aligned} S_{n} & = \frac{a (1 - r^{n})}{1 - r} \\ = \frac{a (r^{n} - 1)}{r - 1} \end{aligned}

(8.3)

3/3Sirf $a$ aur $a r^{n}$ bachte hain: $(1 - r) S_{n} = a - a r^{n}$ .

Fig. 8.7Ek jagah shift karo, subtract karo, aur lagbhag sab kuch cancel ho jaata hai.

Upar ki row mein likha hai S sub n equals a plus ar plus ar squared plus dots plus ar to the n minus 1. Doosri row mein likha hai r S sub n equals ar plus ar squared plus dots plus ar to the n minus 1 plus ar to the n, jo ek jagah right mein likhi hai taaki barabar terms shaded columns mein aa jaayein. Doosre step mein shaded columns ka har matching pair kaat diya jaata hai. Teesre step mein do bache hue terms, upar a aur neeche ar to the n, gol ghere mein hain, aur neeche lines one minus r times S sub n equals a minus ar to the n aur S sub n ka final formula dikhte hain.

(8.3) ke dono forms ek hi fraction hain, bas upar aur neeche dono ko $- 1$ se multiply kiya gaya hai. Pehla form $r < 1$ hone par kaam ka hai aur doosra $r > 1$ hone par, kyunki tab jin numbers ke saath tum kaam karte ho woh positive rehte hain. Exception yaad rakho: $r = 1$ ke liye $S_{n} = n a$ . Neeche wala lab tumhe $a$ , $r$ aur $n$ badalne aur formula ko seedhe jodne se compare karne deta hai.

InteractiveProgression lab

Do modes. Terms and sums mein sliders first term a ko −5 se 5 tak, common ratio r ko quarter steps mein −3 se 3 tak aur terms ki sankhya n ko 1 se 12 tak set karte hain. Bars terms dikhati hain (positive ke liye blue, negative ke liye orange), green dots ka ek optional set partial sums dikhata hai, aur ek panel nth term, jodkar mila sum aur formula se mila sum deta hai, saath mein r = 1 wala case aur a = 0 ya r = 0 wale cases samjhaata hai. Insert means mein sliders do positive numbers aur kitne means insert karne hain (1 se 5) set karte hain; do number lines arithmetic means ko barabar steps mein aur geometric means ko barabar ratios mein dikhati hain, aur ek mean ke liye panel A aur G ko compare karta hai.

Fig. 8.8Terms aur unka sum dekhne ke liye

a

r

aur

n

set karo, ya do numbers ke beech arithmetic aur geometric means insert karo.

Example 8.4NCERT Example 4 · p. 140

10th aur nth term

G.P. $3, 12, 48, \dots$ ka 10th term aur $n$ th term nikaalo.

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

Example 8.5NCERT Example 5 · p. 141

Yeh kaun sa term hai?

G.P. $5, 10, 20, \dots$ ka kaun sa term 5120 hai?

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

Example 8.6NCERT Example 6 · p. 141

Do terms G.P. ko tay kar dete hain

Ek G.P. ka 2nd term 6 hai aur 5th term $- 162$ hai. Iska 9th term nikaalo.

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

Example 8.7NCERT Example 7 · p. 141

Fractional ratio wala sum

$4 + 3 + \frac{9}{4} + \dots$ ke pehle $n$ terms ka sum nikaalo, aur iske pehle 4 terms ka sum bhi.

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

0 aur 1 ke beech ke ratios ki ek sundar picture banti hai. $a = \frac{1}{2}$ aur $r = \frac{1}{2}$ ke saath, (8.3) se $S_{n} = 1 - {(\frac{1}{2})}^{n} .$ milta hai. Fig. 8.9 mein har naya term ek unit square ki bachi hui khaali jagah ka aadha bharta hai, aur jo gap bachta hai woh hamesha aakhri jode gaye piece ke bilkul barabar hota hai.

4/4 $S_{4} = \frac{15}{16}$ ; gap $\frac{1}{16}$ aakhri piece ke barabar hai.

Fig. 8.9Har piece bachi hui jagah ka aadha bharta hai, isliye

S_{n} = 1 - {(\frac{1}{2})}^{n} .

Area 1 wale square ki outline bani hai. Pehle step mein uska left half shaded hai aur one half label hai, aur right side mein S1 equals one half likha hai. Doosre step mein right half ka upar wala hissa quarter ki tarah shaded hai, aur S2 equals three quarters. Teesre step mein bachi hui jagah ka aadha eighth ki tarah shaded hai, aur S3 equals seven eighths. Chauthe step mein ek sixteenth judta hai, ek dashed square one sixteenth ke khaali gap ko mark karta hai, aur text mein S4 equals fifteen sixteenths aur S sub n equals 1 minus one half to the power n likha hai.

Example 8.8NCERT Example 8 · p. 141

Kitne terms?

728 paane ke liye G.P. $2, 6, 18, \dots$ ke kitne terms jodne padenge?

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

Example 8.9NCERT Example 9 · p. 142

Sum aur product se teen terms

Ek G.P. ke teen lagaatar terms ka sum $\frac{19}{2}$ hai aur product 27 hai. Unhe nikaalo.

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

Example 8.10NCERT Example 10 · p. 142

Ek sum jismein G.P. chhupi hai

$2 + 22 + 222 + \dots$ ke pehle $n$ terms ka sum nikaalo

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

Example 8.11NCERT Example 11 · p. 143

Knockout mein matches

Ek badminton knockout tournament 128 players se shuru hota hai. Har match ek player ko bahar karta hai, aur jeetne wale tab tak khelte rehte hain jab tak ek champion na bache. Kitne matches khele jaate hain?

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

4/4 $8 + 4 + 2 + 1 = 15 = 16 - 1$ matches.

Fig. 8.10Knockout ke rounds ratio

\frac{1}{2}

wali G.P. banate hain.

Left mein solah dots players ko dikhate hain. Pehle step mein brackets unhe round 1 ke 8 matches mein jodte hain, aur text mein Round 1: 8 likha hai. Doosre step mein dashed brackets winners ko 4 matches mein jodte hain, aur Round 2: 4 judta hai. Teesre step mein moti brackets round 3 ke 2 matches aur phir final dikhati hain, jo champion tak le jaata hai, aur lines Round 3: 2 aur Final: 1 judti hain. Chauthe step mein list ke neeche ek line kul 15 matches batati hai, jo 16 minus 1 ke barabar hai.

Quick check

4 mein se 1

Inmein se kaun si geometric progressions hain?

5, - 10, 20, - 40, \dots

2, 4, 6, 8, \dots

81, 27, 9, 3, \dots

3, 0, 0, 0, \dots

Saare sahi options chuno

Answer dekho

Answer

(A) $5, - 10, 20, - 40, \dots$
(C) $81, 27, 9, 3, \dots$

A (

- 2

) aur C (

\frac{1}{3}

) mein ratios constant hain. B mein difference constant hai, isliye woh A.P. hai. D G.P. nahi hai kyunki uske terms non-zero hone chahiye;

0 \div 0

ka koi matlab nahi.

8.4.3 Geometric Mean (G.M.)

Jab teen positive numbers $a, G, b$ G.P. mein hon, to beech wala number $\frac{G}{a} = \frac{b}{G},$ satisfy karta hai, yaani $G^{2} = a b$ . Yeh beech wala number $a$ aur $b$ ka geometric mean hai:

a, b > 0 ka G.M.

G = \sqrt{a b}

(8.4)

Jaise, 4 aur 25 ka G.M. $\sqrt{100} = 10$ hai, aur $4, 10, 25$ ka common ratio sach mein $\frac{5}{2}$ hai. Hum $a$ aur $b$ ko positive rakhte hain taaki $a b > 0$ ho aur root real ho, aur hum positive root lete hain.

$a$ aur $b$ ke beech $n$ numbers $G_{1}, \dots, G_{n}$ aise fit karne ke liye ki poori list G.P. ho, terms gino: $n + 2$ terms hain, aur $b$ aakhri hai, isliye $b = a r^{n + 1}$ . Isliye

n G.M.s insert karna

\begin{aligned} r & = {(\frac{b}{a})}^{1 / (n + 1)} \\ G_{k} & = a r^{k} \end{aligned}

(8.5)

jahan

$k$: $1, 2, \dots, n$

Yeh $G_{1}, \dots, G_{n}$ hi $a$ aur $b$ ke beech ke $n$ geometric means hain. Fig. 8.11 inhe arithmetic means se compare karta hai: 2 aur 32 ke beech teen A.M.s 7.5 ke barabar steps mein badhte hain, jabki teen G.M.s 2 ke barabar factors mein badhte hain. Fig. 8.8 ka Insert means mode tumhe doosre numbers try karne deta hai.

3/3Barabar factors: $r^{4} = \frac{32}{2} = 16$ , isliye $r = 2$ .

Fig. 8.11Arithmetic means barabar doori par hote hain; geometric means barabar multiply hote hain.

Do horizontal number lines 2 se 32 tak chalti hain, upar wali ka title Arithmetic means aur neeche wali ka Geometric means hai. Pehle step mein dono par sirf end points 2 aur 32 mark hain. Doosre step mein upar wali line par 9.5, 17 aur 24.5 par points aate hain, aur neeche chaar barabar arcs hain jin par plus 7.5 likha hai. Teesre step mein neeche wali line par 4, 8 aur 16 par points aate hain, aur times 2 label wale arcs left se right lambe hote jaate hain, aakhri arc 16 se poora 32 tak jaata hai.

Example 8.12NCERT Example 12 · p. 144

Teen geometric means insert karna

2 aur 162 ke beech teen numbers insert karo taaki paanchon numbers G.P. banayein.

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

Quick check

3 mein se 1

3 aur 27 ka geometric mean nikaalo.

Answer dekho

Answer 9

\sqrt{81} = 9

; sach mein

3, 9, 27

ka ratio 3 hai. Arithmetic mean, 15, ek alag average hai.

8.5 A.M. aur G.M. ke beech relationship

Do positive numbers $a$ aur $b$ ke liye ab hamare paas do averages hain: arithmetic mean $A = \frac{a + b}{2}$ , jo A.P. ka beech wala term hai, aur geometric mean $G = \sqrt{a b}$ , jo G.P. ka beech wala term hai. Kuch pairs try karo. 2 aur 8 ke liye $A = 5$ aur $G = 4$ ; 9 aur 16 ke liye $A = 12.5$ aur $G = 12$ ; 7 aur 7 ke liye dono 7 hain. Lagta hai A.M. kabhi G.M. se neeche nahi girta.

TheoremA.M.–G.M. inequality

Agar $a$ aur $b$ positive real numbers hain jinka arithmetic mean $A$ aur geometric mean $G$ hai, to $A \geq G$ , aur $A = G$ theek tabhi jab $a = b$ .

a, b > 0 ke liye A.M. ≥ G.M.

\frac{a + b}{2} \geq \sqrt{a b}

(8.6)

Iska ek picture proof bhi hai (Fig. 8.12). Length $a + b$ ke diameter par ek semicircle banao, jo $a$ aur $b$ pieces mein bata ho. Iska radius $A$ hai. Split point se arc tak ke perpendicular ki length $G$ hai: semicircle mein angle right angle hota hai, aur right triangle ke hypotenuse par altitude un do pieces ka G.M. hoti hai jinmein woh use kaatti hai. Koi bhi half-chord radius se lambi nahi hoti, isliye $G \leq A$ .

3/4Right triangle mein altitude $G^{2} = a b$ satisfy karti hai, aur $G < A$ .

Fig. 8.12Radius A.M. hai; split point par half-chord G.M. hai.

Ek shaded semicircle horizontal diameter par khada hai, jise ek tick left mein lambe piece a aur right mein chhote piece b mein baantta hai. Doosre step mein centre se arc ke top tak ek vertical radius uthta hai, jis par A equals a plus b over 2 likha hai. Teesre step mein split point se arc tak ek dashed vertical segment uthta hai, jis par G equals root ab likha hai, aur dashed lines uske top ko diameter ke dono ends se jodti hain, jisse right triangle banta hai; segment saaf taur par radius se chhota hai. Chauthe step mein split point centre tak khisakta hai, yaani a equals b, aur dashed segment lamba hokar theek radius ko cover kar leta hai, jo A equals G dikhata hai.

Example 8.13NCERT Example 13 · p. 144

Means se numbers

Do positive numbers ka A.M. 15 hai aur G.M. 12 hai. Numbers nikaalo.

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

Yeh inequality 'best value' wale sawaal bhi hal karti hai. Maan lo 40 m fencing se sides $x$ aur $y$ wala ek rectangular plot gherna hai, to $x + y = 20$ . (8.6) se $\sqrt{x y} \leq \frac{x + y}{2} = 10,$ isliye area $x y$ zyada se zyada 100 m² hai, aur 100 m² sirf tab hota hai jab $x = y = 10$ . Fixed sum ke liye, product tab sabse bada hota hai jab dono numbers barabar hon.

Quick check

4 mein se 1

Do unequal positive numbers ke liye kaun sa statement hamesha true hai?

A = G

A > G

A < G

Yeh numbers par depend karta hai

Answer dekho

Answer (B) $A > G$

A - G

ek square ka aadha hai, aur square positive hai kyunki

\sqrt{a} \neq \sqrt{b}

. Equality ke liye

a = b

chahiye, jo yahan allowed nahi hai.

Practice · 29 questions

Practice 8.2: answers ke saath

Short answerEasyNCERT Exercise 8.2, Q1
G.P. $\frac{3}{4}, \frac{3}{2}, 3, \dots$ ka 20th term aur $n$ th term nikaalo
Solution dikhao
$a = \frac{3}{4}$ aur $r = 2$ , isliye $a_{n} = \frac{3}{4} \cdot 2^{n - 1} = 3 \cdot 2^{n - 3} .$
Phir $a_{20} = 3 \cdot 2^{17} = 393 216.$
Quick answerEasyNCERT Exercise 8.2, Q2
Common ratio 2 wali G.P. mein 10th term 1536 hai. 13th term nikaalo.
Answer dekho
Answer 12288
$a_{13} = a_{10} r^{3} = 1536 \times 8 = 12 288.$ Tumhe $a$ ki zaroorat hi nahi padti (woh 3 hai).
ProofMediumNCERT Exercise 8.2, Q3
Ek G.P. mein 3rd term ko $p$ , 9th term ko $q$ aur 15th term ko $s$ kaho. Dikhao ki $q^{2} = p s$ .
Solution dikhao
$p = a r^{2}$ , $q = a r^{8}$ aur $s = a r^{14}$ .
To $p s = a^{2} r^{16} = (a r^{8})^{2} = q^{2} .$
Short answerMediumNCERT Exercise 8.2, Q4
Ek G.P. $- 2$ se shuru hoti hai. Iske 3rd term ka square karne par 6th term milta hai. 8th term nikaalo.
Solution dikhao
$a r^{5} = a^{2} r^{4}$ . Terms non-zero hain, isliye $a r^{4}$ se divide karo: $r = a = - 2$ .
Phir $a_{8} = a r^{7} = (- 2) (- 2)^{7} = 256.$ Check: $a_{3} = - 8$ aur $a_{6} = 64 = (- 8)^{2}$ .
5
Quick answerMediumNCERT Exercise 8.2, Q5
Har G.P. ka kaun sa term diya gaya number hai?
1. $\sqrt{2}, 2, 2 \sqrt{2}, \dots$ mein 64
  Answer dekho
  Answer 12
  Yahan $a = r = \sqrt{2}$ , isliye $a_{n} = (\sqrt{2})^{n} = 2^{n / 2} .$ Phir $2^{n / 2} = 2^{6}$ se $n = 12$ milta hai.
2. $\sqrt{5}, 5, 5 \sqrt{5}, \dots$ mein 3125
  Answer dekho
  Answer 10
  $a_{n} = (\sqrt{5})^{n} = 5^{n / 2}$ aur $3125 = 5^{5}$ , isliye $n = 10$ .
3. $\frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \dots$ mein $\frac{2}{6561}$
  Answer dekho
  Answer 8
  $a_{n} = \frac{2}{3^{n}}$ aur $6561 = 3^{8}$ , isliye $n = 8$ .
Short answerEasyNCERT Exercise 8.2, Q6
$x$ ki kin values ke liye $- \frac{3}{5}, x, - \frac{20}{3}$ G.P. mein hain?
Solution dikhao
$x^{2} = (- \frac{3}{5}) (- \frac{20}{3}) = 4,$ isliye $x = 2$ ya $x = - 2$ .
Dono chalte hain: ratios $- \frac{10}{3}$ aur $\frac{10}{3}$ hain, usi order mein.
7
Short answerMediumNCERT Exercise 8.2, Q7–Q10
Har sum nikaalo.
1. $0.4 + 0.04 + 0.004 + \dots$ ke 15 terms tak
  Solution dikhao
  $a = 0.4$ , $r = 0.1$ : $S_{15} = \frac{0.4 (1 - {0.1}^{15})}{0.9} = \frac{4}{9} (1 - 10^{- 15}) .$
2. $\sqrt{2} + \sqrt{6} + 3 \sqrt{2} + \dots$ ke $n$ terms tak
  Solution dikhao
  $a = \sqrt{2}$ , $r = \sqrt{3}$ : $S_{n} = \frac{\sqrt{2} (3^{n / 2} - 1)}{\sqrt{3} - 1} .$
  Rationalise karne par, $S_{n} = \frac{\sqrt{6} + \sqrt{2}}{2} (3^{n / 2} - 1) .$
3. $1 - 2 x + 4 x^{2} - 8 x^{3} + \dots$ ke $n$ terms tak ( $x \neq - \frac{1}{2}$ )
  Solution dikhao
  $a = 1$ , $r = - 2 x \neq 1$ : $S_{n} = \frac{1 - (- 2 x)^{n}}{1 + 2 x} .$
4. $y^{2} + y^{5} + y^{8} + \dots$ ke $n$ terms tak ( $y \neq 1$ )
  Solution dikhao
  $a = y^{2}$ , $r = y^{3} \neq 1$ : $S_{n} = \frac{y^{2} (y^{3 n} - 1)}{y^{3} - 1} .$
Quick answerMediumNCERT Exercise 8.2, Q11
$\sum_{k = 1}^{8} (5 - 2^{k})$ evaluate karo.
Answer dekho
Answer -470
$\sum 5 = 40$ aur $2 + 4 + \dots + 256 = 2 (2^{8} - 1) = 510,$ isliye sum $40 - 510 = - 470$ hai.
Short answerMediumNCERT Exercise 8.2, Q12
G.P. mein teen aise numbers nikaalo jinka product 8 aur sum 7 ho. Common ratio kya hai?
Solution dikhao
Product $a^{3} = 8$ hai, isliye $a = 2$ .
Phir $\frac{2}{r} + 2 + 2 r = 7$ , yaani $2 r^{2} - 5 r + 2 = 0$ , isliye $r = 2$ ya $r = \frac{1}{2}$ .
Numbers 1, 2, 4 (ratio 2) ya 4, 2, 1 (ratio $\frac{1}{2}$ ) hain.
Quick answerEasyNCERT Exercise 8.2, Q13
4 se shuru karke baari-baari $4^{2}$ , $4^{3}$ , … jodte hue, theek 1364 tak pahunchne ke liye 4 ki kitni powers jodni padengi?
Answer dekho
Answer 5
$\frac{4 (4^{n} - 1)}{3} = 1364$ se $4^{n} - 1 = 1023$ milta hai, isliye $4^{n} = 1024 = 4^{5}$ aur $n = 5$ .
Short answerMediumNCERT Exercise 8.2, Q14
Ek G.P. ke 1st, 2nd aur 3rd terms ka total 7 hai, jabki 4th, 5th aur 6th ka total 56 hai. $a$ , $r$ aur $S_{n}$ ka formula nikaalo.
Solution dikhao
$a (1 + r + r^{2}) = 7$ aur $a r^{3} (1 + r + r^{2}) = 56$ . Divide karne par $r^{3} = 8$ , isliye $r = 2$ aur $a = 1$ .
Phir $S_{n} = \frac{2^{n} - 1}{2 - 1} = 2^{n} - 1.$
Quick answerMediumNCERT Exercise 8.2, Q15
Ek G.P. ka 6th term 243 hai aur first term 1024 hai. Iske pehle chhah terms jodo.
Answer dekho
Answer 3367
$r^{5} = \frac{3^{5}}{4^{5}}$ , isliye $r = \frac{3}{4}$ . $S_{6} = \frac{1024 [1 - (\frac{3}{4})^{6}]}{\frac{1}{4}} = 4096 - 729 = 3367.$
Short answerMediumNCERT Exercise 8.2, Q16
Ek G.P. ka 4th term uske 2nd term ka nau guna hai, aur pehle do terms ka total 12 hai. Aisi har G.P. nikaalo.
Solution dikhao
$a r^{3} = 9 a r$ , isliye $r^{2} = 9$ aur $r = \pm 3$ .
Agar $r = 3$ : $a (1 + 3) = 12$ , isliye $a = 3$ aur G.P. $3, 9, 27, \dots$ hai
Agar $r = - 3$ : $a (1 - 3) = 12$ , isliye $a = - 6$ aur G.P. $- 6, 18, - 54, \dots$ hai
ProofMediumNCERT Exercise 8.2, Q17
Maan lo $x$ , $y$ aur $z$ ek G.P. ke 2nd, 7th aur 12th terms hain. Prove karo ki $x, y, z$ bhi G.P. banate hain.
Solution dikhao
$x = a r$ , $y = a r^{6}$ , $z = a r^{11}$ , isliye $x z = a^{2} r^{12} = y^{2}$ , aur teenon non-zero hain. Isliye $x, y, z$ ratio $r^{5}$ ke saath G.P. mein hain.
Yahi tab bhi kaam karta hai jab positions A.P. banati hain (yahan 2, 7, 12): position mein barabar gaps se $r$ ki barabar powers milti hain.
Short answerMediumNCERT Exercise 8.2, Q18
$3 + 33 + 333 + \dots$ ke pehle $n$ terms ka sum nikaalo
Solution dikhao
$S_{n} = \frac{3}{9} [(10 + 10^{2} + \dots + 10^{n}) - n] .$
$S_{n} = \frac{1}{3} [\frac{10 (10^{n} - 1)}{9} - n] = \frac{10}{27} (10^{n} - 1) - \frac{n}{3} .$
Quick answerEasyNCERT Exercise 8.2, Q19
$3, 6, 12, 24$ aur $64, 16, 4, 1$ ke corresponding terms ko multiply karo, aur chaaron products jodo.
Answer dekho
Answer 360
Products $192, 96, 48, 24$ hain, jo ratio $\frac{1}{2}$ wali G.P. hai; inka sum 360 hai.
ProofMediumNCERT Exercise 8.2, Q20
Dikhao ki G.P. $a, a r, a r^{2}, \dots$ ke terms ko G.P. $A, A R, A R^{2}, \dots$ ke corresponding terms se divide karne par ek G.P. milti hai, aur uska common ratio nikaalo.
Solution dikhao
$k$ th quotient $\frac{a r^{k - 1}}{A R^{k - 1}} = \frac{a}{A} {(\frac{r}{R})}^{k - 1} .$ hai
Yeh G.P. hai jiska first term $\frac{a}{A}$ aur common ratio $\frac{r}{R}$ hai (saare terms non-zero).
Short answerMediumNCERT Exercise 8.2, Q21
G.P. mein aise chaar numbers nikaalo ki 3rd, 1st se 40 zyada ho aur 4th, 2nd se 120 zyada ho.
Solution dikhao
$a (r^{2} - 1) = 40$ aur $a r (r^{2} - 1) = 120$ . Divide karne par $r = 3$ ; phir $8 a = 40$ , isliye $a = 5$ .
Numbers 5, 15, 45, 135 hain.
ProofHardNCERT Exercise 8.2, Q22
Positive terms wali ek G.P. ka first term $A$ aur ratio $R$ hai. Positions $p$ , $q$ aur $r$ par iske terms $x$ , $y$ aur $z$ hain. Prove karo ki $y^{r - p} = x^{r - q} z^{q - p} .$
Solution dikhao
$x = A R^{p - 1}$ , $y = A R^{q - 1}$ , $z = A R^{r - 1}$ .
$A$ ki powers: left $r - p$ ; right $(r - q) + (q - p) = r - p$ .
$R$ ki powers: left $(q - 1) (r - p)$ ; right $(p - 1) (r - q) + (r - 1) (q - p),$ jo expand karne par $q r - p q - r + p = (q - 1) (r - p) .$ banta hai. Dono sides match karti hain.
20
ProofHardNCERT Exercise 8.2, Q23
Ek G.P. mein $n$ positive terms hain, first term $a$ aur last term $l$ hai.
1. Dikhao ki shuru se $k$ th term aur end se $k$ th term ka product hamesha $a l$ hota hai.
  Solution dikhao
  Woh $a r^{k - 1}$ aur $l r^{- (k - 1)}$ hain; unka product $a l$ hai.
  To, saare terms ke product $P$ ko aage se aur peeche se likhkar aur dono lines ko term by term multiply karke, $P^{2} = (a l)^{n}$ .
2. Isse $2, 6, 18, \dots, 486$ ke chhah terms ka product 972 ki power ke roop mein nikaalo.
  Solution dikhao
  $a = 2$ , $l = 486$ , $n = 6$ : $P^{2} = 972^{6}$ , isliye $P = 972^{3}$ (terms positive hain). Check: product $2^{6} \cdot 3^{15} = (2^{2} \cdot 3^{5})^{3} .$ hai
ProofMediumNCERT Exercise 8.2, Q24
Ek G.P. ke liye, $S_{1}$ pehle $n$ terms ka sum, $S_{2}$ agle $n$ terms ka sum aur $S_{3}$ unke baad ke $n$ terms ka sum hai. Prove karo ki $S_{2}^{2} = S_{1} S_{3}$ .
Solution dikhao
Doosre block ka har term pehle block ke matching term ka $r^{n}$ guna hai, isliye $S_{2} = r^{n} S_{1}$ ; isi tarah $S_{3} = r^{2 n} S_{1}$ .
Isliye $S_{1} S_{3} = r^{2 n} S_{1}^{2} = S_{2}^{2} .$
ProofMediumNCERT Exercise 8.2, Q25
Agar $a, b, c$ G.P. mein hain, to prove karo ki $(a + b + c) (a - b + c) = a^{2} + b^{2} + c^{2} .$
Solution dikhao
Left side $(a + c)^{2} - b^{2} = a^{2} + 2 a c + c^{2} - b^{2} .$ hai
$a c = b^{2}$ ke saath yeh $a^{2} + 2 b^{2} + c^{2} - b^{2} = a^{2} + b^{2} + c^{2} .$ hai
Short answerEasyNCERT Exercise 8.2, Q26
5 aur 320 ke beech do numbers insert karo taaki chaaron numbers G.P. mein hon.
Solution dikhao
$r^{3} = 64$ , isliye $r = 4$ , aur numbers 20 aur 80 hain.
Short answerHardNCERT Exercise 8.2, Q27
Positive numbers $a \neq b$ ke liye $n$ aisa nikaalo ki $\frac{a^{n} + b^{n}}{a^{n - 1} + b^{n - 1}}$ , $a$ aur $b$ ka G.M. ho.
Solution dikhao
Hamein $m = n - 1$ ke saath $a^{m + 1} + b^{m + 1} = \sqrt{a b} (a^{m} + b^{m})$ chahiye.
Rearrange karo: $a^{m + \frac{1}{2}} (\sqrt{a} - \sqrt{b}) - b^{m + \frac{1}{2}} (\sqrt{a} - \sqrt{b}) = 0.$
Kyunki $a \neq b$ , $a^{m + \frac{1}{2}} = b^{m + \frac{1}{2}},$ jisse $m + \frac{1}{2} = 0$ hona padta hai. Isliye $m = - \frac{1}{2}$ aur $n = \frac{1}{2}$ .
ProofHardNCERT Exercise 8.2, Q28
Do positive numbers ka sum unke geometric mean ka chaar guna hai. Prove karo ki bada number chhote se utne hi ratio mein hai jitna $(2 + \sqrt{3})$ , $(2 - \sqrt{3})$ se.
Solution dikhao
$a + b = 4 \sqrt{a b}$ ban jaata hai $t^{2} - 4 t + 1 = 0$ , isliye $t = 2 \pm \sqrt{3}$ aur $\frac{a}{b} = t^{2} = (2 \pm \sqrt{3})^{2} .$
Kyunki $(2 + \sqrt{3}) (2 - \sqrt{3}) = 1,$ $(2 + \sqrt{3})^{2} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} .$ Isliye $a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}),$ ya ulta.
26
Short answerMediumNCERT Exercise 8.2, Q29
Do positive numbers ka A.M. $A$ aur G.M. $G$ hai.
1. Dikhao ki woh $x^{2} - 2 A x + G^{2} = 0$ ke roots hain.
  Solution dikhao
  Unka sum $2 A$ hai aur product $G^{2}$ hai, isliye woh $x^{2} - 2 A x + G^{2} = 0$ ke roots hain.
2. $A = 17$ aur $G = 8$ hone par numbers nikaalo.
  Solution dikhao
  $x^{2} - 34 x + 64 = 0$ se $x = 17 \pm \sqrt{289 - 64} = 17 \pm 15,$ milta hai, isliye numbers 32 aur 2 hain.
  General mein formula do numbers ke liye $A \pm \sqrt{(A + G) (A - G)}$ deta hai.
Short answerEasyNCERT Exercise 8.2, Q30
Ek cell culture har ghante teen guna hota hai aur 50 cells se shuru hota hai. 3 ghante baad, 5 ghante baad aur $n$ ghante baad kitne cells honge?
Solution dikhao
$n$ ghante baad: $50 \cdot 3^{n}$ cells.
3 ghante baad: 1350; 5 ghante baad: 12 150.
NumericalEasyNCERT Exercise 8.2, Q31
₹20,000, 8% saalana par jama kiye jaate hain, har saal compound hote hue. 5 saal baad amount kitna hai, rupees mein?
Answer dekho
Answer 29386.56
Amounts ratio 1.08 wali G.P. banate hain, isliye 5 saal baad amount $20 000 \times {1.08}^{5} \approx$ ₹29,386.56 hai.
Short answerMediumNCERT Exercise 8.2, Q32
Ek quadratic equation $x^{2} + p x + q = 0$ ke positive roots hain jinka arithmetic mean 13 aur geometric mean 5 hai. $p$ aur $q$ nikaalo.
Solution dikhao
Roots ka sum $= 26 = - p$ aur product $= 25 = q$ , isliye $p = - 26$ , $q = 25$ : equation $x^{2} - 26 x + 25 = 0$ hai, jiske roots 1 aur 25 hain.

Miscellaneous Examples

Example 8.14NCERT Example 14 · p. 147

Jab squares ka sum positive na ho

Maan lo $a, b, c$ non-zero real numbers hain, aur koi real $x$ $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) \leq 0.$ ko satisfy karta hai. Dikhao ki $a, b, c$ common ratio $x$ ke saath G.P. mein hain.

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

Exercises mein roz-marra ke do tarah ke problems seedhe progressions use karte hain. Instalment plan mein har saal ka interest abhi tak na chukaaye gaye amount par lagta hai, aur woh amount har saal ek fixed instalment se ghatta hai, isliye interest payments ek A.P. banate hain.

Example

Instalments mein payment

Meera ₹60,000 ka ek laptop kharidti hai. Woh ₹20,000 turant deti hai aur baaki ₹40,000 ₹8,000 ki yearly instalments mein chukaati hai, har instalment ke saath us saal ki shuruaat mein bache unpaid amount par 10% interest bhi deti hai. Laptop use kul kitne ka padta hai?

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

Interest dono progressions ko saath-saath dikhata hai. Simple interest par ₹10,000, 10% saalana par har saal same ₹1,000 kamaata hai, isliye har saal ke end ke amounts A.P. banate hain. Compound interest par har saal ka interest principal mein jud jaata hai, isliye amount har saal 1.1 se multiply hota hai: ₹11,000, ₹12,100, ₹13,310, …, yaani G.P. Depreciation bhi isi tarah kaam karta hai, bas ratio 1 se kam hota hai: jo machine har saal apni value ka 10% khoti hai, woh uska 0.9 bachaati hai (Fig. 8.13).

3/35 saal baad: ₹15,000 ke saamne ₹16,105, aur gap badhta hi jaata hai.

Fig. 8.13Simple interest ek fixed amount jodta hai (A.P.); compound interest ek fixed factor se multiply karta hai (G.P.).

Ek graph jismein across years 0 se 5 aur side mein amounts 10,000 se 16,000 rupees tak hain. Pehle step mein dashed straight line se jude square markers simple interest dikhate hain: 10,000, 11,000, 12,000 aur aise hi 15,000 tak, har saal 1,000 badhte hue. Doosre step mein solid line se jude round markers compound interest dikhate hain: 10,000, 11,000, 12,100, 13,310, 14,641 aur 16,105, upar ki taraf mudte hue. Teesre step mein year 5 par ek chhoti bar 15,000 aur 16,105 ke beech ka gap mark karti hai.

Quick check

2 mein se 1

₹5,000, 10% saalana par badhta hai, har saal compound hote hue. 2 saal baad iski value kitni hai, rupees mein?

Answer dekho

Answer 6050

5000 \times {1.1}^{2} = 6050

. Simple interest se sirf 6000 milte, kyunki usmein interest par interest kabhi nahi milta.

Recap

Summary aur key points

Sequence ek pakke order wali list hai; formally $𝐍$ ya ${1, \dots, k}$ par ek function. Yeh finite ya infinite hota hai (Section 8.2).
Terms $a_{n}$ ke formula se, starting terms wale recurrence relation se (Fibonacci), ya shabdon mein diye rule se ( $π$ ke digits) aate hain.
Series expression $a_{1} + a_{2} + \dots$ hai; uska sum woh number hai jo milta hai. $\sum_{k = 1}^{n} a_{k}$ sigma notation hai (Section 8.3).
G.P. ke terms non-zero hote hain aur unka ratio $r = \frac{a_{k + 1}}{a_{k}}$ constant hota hai (Section 8.4).
$n$ th term $a_{n} = a r^{n - 1}$ hai ((8.2)).
$r \neq 1$ ke liye $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ , aur $r = 1$ ke liye $S_{n} = n a$ ((8.3)).
Kaam ke set-ups: teen terms ke liye $\frac{a}{r}, a, a r$ ; $7 + 77 + \dots$ ke liye $(10^{k} - 1)$ (Section 8.4.2).
Positive $a$ aur $b$ ka G.M. $\sqrt{a b}$ hai; $n$ G.M.s ke liye $r = (b / a)^{1 / (n + 1)}$ use hota hai (Section 8.4.3).
Positive numbers ke liye $A \geq G$ , kyunki $A - G = \frac{1}{2} (\sqrt{a} - \sqrt{b})^{2};$ equality sirf tab jab $a = b$ (Section 8.5).
Compound interest aur depreciation G.P.s hain; simple interest aur instalments A.P.s hain (Miscellaneous Examples).

Timeline

Progressions ko kaise study kiya gaya

c. 1800 BCE
Babylonian tablets
Scribes aisi lists ke saath kaam karte the jo jodne aur double karne se badhti hain, jinmein badhte loans ke problems bhi the.
c. 300 BCE
Euclid
Elements ki Book IX geometric progression ko jodne ka ek rule deti hai.
499 CE
Aryabhata
Aryabhatiya A.P. ka sum aur natural numbers ke squares aur cubes ke sums deta hai.
c. 510
Boethius
Unki arithmetic ki textbook progressions aur means par Greek kaam ko medieval Europe tak le jaati hai.
c. 850
Mahavira
Ganitasarasangraha geometric progressions ke sums ke rules deta hai; Brahmagupta aur baad mein Bhaskara II ne bhi squares aur cubes ke sums ko study kiya.
c. 1150
Hemachandra
Chhote aur lambe syllables wale poetic rhythms ginte hain; yeh counts Fibonacci rule follow karte hain.
1202
Fibonacci
Liber Abaci mein rabbits ke badhne ki ek puzzle hai jisse 1, 1, 2, 3, 5, 8, … milta hai
c. 1400
Madhava
Kerala ke mathematicians π aur sine aur cosine ke liye infinite series dhoondhte hain.
1671
James Gregory
Arctangent ki infinite series dhoondhte hain; infinite series analysis ka ek central tool ban jaati 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 8.1, Q1–Q3	Practice 8.1 · Q1
NCERT Exercise 8.1, Q4–Q6	Practice 8.1 · Q2
NCERT Exercise 8.1, Q7	Practice 8.1 · Q3
NCERT Exercise 8.1, Q8	Practice 8.1 · Q4
NCERT Exercise 8.1, Q9	Practice 8.1 · Q5
NCERT Exercise 8.1, Q10	Practice 8.1 · Q6
NCERT Exercise 8.1, Q11–Q13	Practice 8.1 · Q7
NCERT Exercise 8.1, Q14	Practice 8.1 · Q8
NCERT Exercise 8.2, Q1	Practice 8.2 · Q1
NCERT Exercise 8.2, Q2	Practice 8.2 · Q2
NCERT Exercise 8.2, Q3	Practice 8.2 · Q3
NCERT Exercise 8.2, Q4	Practice 8.2 · Q4
NCERT Exercise 8.2, Q5	Practice 8.2 · Q5
NCERT Exercise 8.2, Q6	Practice 8.2 · Q6
NCERT Exercise 8.2, Q7–Q10	Practice 8.2 · Q7
NCERT Exercise 8.2, Q11	Practice 8.2 · Q8
NCERT Exercise 8.2, Q12	Practice 8.2 · Q9
NCERT Exercise 8.2, Q13	Practice 8.2 · Q10
NCERT Exercise 8.2, Q14	Practice 8.2 · Q11
NCERT Exercise 8.2, Q15	Practice 8.2 · Q12
NCERT Exercise 8.2, Q16	Practice 8.2 · Q13
NCERT Exercise 8.2, Q17	Practice 8.2 · Q14
NCERT Exercise 8.2, Q18	Practice 8.2 · Q15
NCERT Exercise 8.2, Q19	Practice 8.2 · Q16
NCERT Exercise 8.2, Q20	Practice 8.2 · Q17
NCERT Exercise 8.2, Q21	Practice 8.2 · Q18
NCERT Exercise 8.2, Q22	Practice 8.2 · Q19
NCERT Exercise 8.2, Q23	Practice 8.2 · Q20
NCERT Exercise 8.2, Q24	Practice 8.2 · Q21
NCERT Exercise 8.2, Q25	Practice 8.2 · Q22
NCERT Exercise 8.2, Q26	Practice 8.2 · Q23
NCERT Exercise 8.2, Q27	Practice 8.2 · Q24
NCERT Exercise 8.2, Q28	Practice 8.2 · Q25
NCERT Exercise 8.2, Q29	Practice 8.2 · Q26
NCERT Exercise 8.2, Q30	Practice 8.2 · Q27
NCERT Exercise 8.2, Q31	Practice 8.2 · Q28
NCERT Exercise 8.2, Q32	Practice 8.2 · Q29
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14	Miscellaneous exercises · Q14
NCERT Miscellaneous Exercise, Q15	Miscellaneous exercises · Q15
NCERT Miscellaneous Exercise, Q16	Miscellaneous exercises · Q16
NCERT Miscellaneous Exercise, Q17	Miscellaneous exercises · Q17
NCERT Miscellaneous Exercise, Q18	Miscellaneous exercises · Q18

Quick check: MCQs

8.2 Sequences4 questions
8.3 Series3 questions
8.4.2 G.P. ke n terms ka sum4 questions
8.4.3 Geometric Mean (G.M.)3 questions
8.5 A.M. aur G.M. ke beech relationship4 questions
Miscellaneous Examples2 questions

Glossary · 13

Key terms (English + Hindi)

arithmetic meanसमांतर माध्यsamantar madhya: Numbers $a$ aur $b$ ke liye number $A = \frac{a + b}{2}$ ; tab $a, A, b$ ek A.P. hai.Chapter mein pehli baar yahan aaya →
arithmetic progressionसमांतर श्रेढ़ीsamantar shredhi: Aisa sequence jismein har term pichhla term plus ek fixed common difference $d$ hota hai.Chapter mein pehli baar yahan aaya →

common ratioसार्व अनुपातsarv anupat: G.P. ka constant $r = \frac{a_{k + 1}}{a_{k}}$ ; yeh kabhi 0 nahi hota.Chapter mein pehli baar yahan aaya →

Fibonacci sequence: Sequence $1, 1, 2, 3, 5, 8, \dots$ jismein second ke baad ka har term apne pichhle do terms ka sum hai.Chapter mein pehli baar yahan aaya →

general termव्यापक पदvyapak pad: Position $n$ ke terms mein $a_{n}$ ka formula, jaise $a_{n} = 4 n - 1$ .Chapter mein pehli baar yahan aaya →
geometric meanगुणोत्तर माध्यgunottar madhya: Positive $a$ aur $b$ ke liye number $G = \sqrt{a b}$ ; tab $a, G, b$ ek G.P. hai.Chapter mein pehli baar yahan aaya →
geometric progressionगुणोत्तर श्रेढ़ीgunottar shredhi: Non-zero numbers ka aisa sequence jismein har term pichhla term times ek fixed common ratio $r$ hota hai.
geometric seriesगुणोत्तर श्रेणीgunottar shreni: G.P. se bani series $a + a r + a r^{2} + \dots$ .Chapter mein pehli baar yahan aaya →

recurrence relation: Aisa rule jo har term ko pehle ke terms se banata hai, first term ya terms ke saath diya jaata hai.Chapter mein pehli baar yahan aaya →

sequenceअनुक्रमanukram: Ek pakke order mein numbers ki list; formally $𝐍$ ya ${1, 2, \dots, k}$ par defined ek function.Chapter mein pehli baar yahan aaya →
seriesश्रेणीshreni: Sequence ke terms jodkar bana expression $a_{1} + a_{2} + a_{3} + \dots$ ; iska sum woh number hai jo milta hai.Chapter mein pehli baar yahan aaya →
sigma notation: Sum ka short form: $\sum_{k = 1}^{n} a_{k}$ ka matlab $a_{1} + a_{2} + \dots + a_{n}$ hai.Chapter mein pehli baar yahan aaya →

termपदpad: Sequence ka ek number; $a_{n}$ position $n$ wala term hai.Chapter mein pehli baar yahan aaya →

Sequences and Series: Class 11 Maths Chapter 8 Hinglish mein

Is chapter mein aap kya seekhenge

8.1 Introduction

8.2 Sequences

8.3 Series

Practice 8.1: answers ke saath

8.4 Geometric Progression (G.P.)

8.4.1 G.P. ka general term

8.4.2 G.P. ke n terms ka sum

8.4.3 Geometric Mean (G.M.)

8.5 A.M. aur G.M. ke beech relationship

Practice 8.2: answers ke saath

Miscellaneous Examples

Summary aur key points

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye