Arithmetic Progressions: Class 10 Maths Chapter 5 Hinglish mein

Barabar steps mein badhti lists: koi bhi term, aur bahut saare terms ka sum, ek line mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
61 practice questions
16 MCQs

Is chapter mein aap kya seekhenge

Neighbouring terms ke beech har difference same hai ya nahi, yeh check karke tay karo ki numbers ki koi list arithmetic progression hai ya nahi, aur uska first term aur common difference pehchaano
First term aur common difference se AP likho, general form a, a + d, a + 2d, … use karo, aur finite APs ko infinite APs se alag pehchaano
nth term a + (n − 1)d derive karo aur usse koi term nikaalo, kisi term ki position nikaalo, ya dikhao ki koi number term nahi hai
Do terms se AP nikaalo, finite AP ke terms gino aur end se gina gaya term nikaalo
Sum ko aage se aur peeche se likh kar first n terms ka sum, n/2 [2a + (n − 1)d] = n/2 (a + l), derive karo
nth term ko do sums ke difference ki tarah use karo, aur samjhao ki number of terms wale sawaal ke do answers kyun ho sakte hain
Roz ki situations, jaise rows mein seats, savings, instalments aur simple interest, ko APs ki tarah model karo aur solve karo

NCERT Mathematics Class 10, Chapter 5, pp. 49–72 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

5.1 Introduction

Matchsticks ko aise rakho ki ek row mein squares banein, aur har naya square pichhle square ke saath ek side share kare. Ek square ke liye 4 sticks chahiye, do squares ke liye 7 aur teen ke liye 10. Har naya square theek 3 sticks zyada leta hai. Aisi lists jo har step par same amount se badalti hain, hamare chaaron taraf hain: fares, rents, savings plans, seating plans aur loan instalments mein.

Har list aisi nahi hoti. Yeh rahi paanch lists, har ek ka apna rule hai:

Ek dukaan ka monthly rent (₹ mein) saal 1, 2, 3 aur 4 mein, jab woh ₹12,000 se shuru hota hai aur har saal ₹600 badhta hai: 12000, 12600, 13200, 13800, …
Ek row mein 1, 2, 3 aur 4 squares ke liye chahiye matchsticks: 4, 7, 10, 13, …
Har bounce ke baad ball jitni height (cm mein) tak jaati hai, jab har bounce pichhle se aadha ooncha ho: 160, 80, 40, 20, …
Handshakes, jab 2, 3, 4 aur 5 dost sab ek doosre se haath milaate hain: 1, 3, 6, 10, …
Class 9 ke Virahānka numbers, jismein har number apne pehle ke do numbers ka sum hai: 1, 2, 3, 5, 8, …

Rent aur matchsticks har baar ek fixed number add karke badhte hain. Ball ki heights har baar $\frac{1}{2}$ se multiply hoti hain. Handshake wali list mein jumps khud badhte hain: 2, phir 3, phir 4. Is chapter mein hum sirf pehli tarah ki list padhenge. Hum koi bhi term nikaalne ka ek quick tareeka dhoondhenge, bahut saare terms ko ek saath jodne ka bhi ek quick tareeka dhoondhenge, aur dono ko roz ke problems par use karenge.

3/3Double karne se jumps 50, 100 aur 200 aate hain: woh badhte rehte hain, isliye jumps barabar nahi hain.

Fig. 5.1Fixed number add karne se barabar jumps bante hain; double karne se nahi.NCERT Fig. 5.1–5.3 · p. 49

Step 1 mein matchstick squares ke chaar groups aamne-saamne hain: ek row mein ek, do, teen aur chaar squares, jin par unke stick counts 4, 7, 10 aur 13 likhe hain. Step 2 counts 4, 7, 10, 13 ko line mein rakhta hai aur har pair ke beech plus 3 label wala arrow banata hai, yaani har jump same hai. Step 3 neeche ek doosri list jodta hai, 50, 100, 200, 400, jismein har number pichhle ka double hai; uske jumps par plus 50, plus 100 aur plus 200 likha hai, jo barabar nahi hain.

Quick check

2 mein se 1

Kaun si list har baar same number add karke badhti hai?

2, 6, 18, 54, …5, 9, 13, 17, …1, 4, 9, 16, …1, 3, 6, 10, …

Answer dekho

Answer (B) 5, 9, 13, 17, …

5, 9, 13, 17 mein har jump 4 hai. List 2, 6, 18, 54 mein har baar 3 se multiply hota hai; squares 1, 4, 9, 16 ke jumps 3, 5, 7 hain; aur 1, 3, 6, 10 ke jumps 2, 3, 4 hain.

5.2 Arithmetic Progressions

Numbers ki in paanch lists ko dekho:

2, 5, 8, 11, …
60, 45, 30, 15, …
−7, −5, −3, −1, …
9, 9, 9, 9, …
1.5, 1.25, 1, 0.75, …

List ke har number ko hum term kehte hain. In sabhi lists mein agla term pichhle term mein same number add karke milta hai: (1) mein 3 add karo, (2) mein −15, (3) mein 2, (4) mein 0 aur (5) mein −0.25. Aisi lists ka ek naam hai.

Definition

Arithmetic progression (AP)

Numbers ki aisi list jismein kisi bhi term se agle term tak same fixed number add karke pahunchte hain (समांतर श्रेढ़ी).

Example5, 9, 13, 17, … (har baar 4 add karo).

Definition

Common difference

Woh amount jitna AP har term se agle term tak badalti hai; ise likhte hain $d .$ AP upar ja sakti hai ( $d > 0$ ), neeche ja sakti hai ( $d < 0$ ) ya same level par reh sakti hai ( $d = 0$ ).

Example60, 45, 30, 15, … mein common difference hai $d = - 15.$

Hum terms ke naam rakhte hain $a_{1}, a_{2}, a_{3}, \dots$ Chhota number term ki position hai, multiplier nahi: $a_{7}$ 7th term hai. AP mein har difference "baad wala term minus pehle wala term" same hota hai, isliye $a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = d .$ Short mein:

Common difference

d = a_{k + 1} - a_{k}

(5.1)

jahan

$a_{k}$: $k$ th term
$a_{k + 1}$: uske theek baad wala term

APs roz ki zindagi mein milti hain. Neeche ki har list ek khaas term ke baad ruk jaati hai. Kya tum dekh sakte ho ki har list AP kyun hai, aur uska $d$ kya hai?

Ek toy xylophone ki bars ki lengths (cm mein), har bar pichhli se 1.5 cm chhoti: 24, 22.5, 21, …, 12
₹6000 ke interest-free loan par abhi kitna baaki hai (₹ mein), jab har mahine ₹500 chukaaye jaate hain: 5500, 5000, 4500, …, 0
December ke ek hafte mein Shimla ke raat ke sabse kam temperatures (°C mein), badhte order mein: −2.4, −2.1, −1.8, …, −0.6
Ek hafte tak water meter ki shaam ki readings (litres mein), jab family roz 450 litres use karti hai: 12350, 12800, …, 15050
Street ki odd side par house numbers: 1, 3, 5, …, 59

$a$ se shuru karo aur $d$ jodte jao: pehle $a,$ phir $a + d,$ phir $a + d + d,$ aur aise hi aage. Yeh AP ka general form hai:

AP ka general form

a, a + d, a + 2 d, \dots

(5.2)

Kuch APs ruk jaati hain aur kuch kabhi nahi rukti. Upar ki roz wali lists mein se har ek kisi fixed number par khatam hoti hai, jaise 12 ya 59, isliye har ek finite AP hai jiska ek last term hai. Is section ki shuru wali paanch lists hamesha chalti rehti hain: har ek infinite AP hai, jiska koi last term nahi.

Do APs ka first term same ho sakta hai aur phir bhi woh alag ho sakti hain: 4, 9, 14, … aur 4, −1, −6, … dono 4 se shuru hoti hain. Do APs ka common difference same hone par bhi woh alag ho sakti hain. AP tabhi poori tarah tay hoti hai jab tumhe dono pata hon, $a$ aur $d,$ jaisa table dikhata hai.

Table 5.1Same first term aur alag-alag common differences se alag-alag APs banti hain.

$a$	$d$	Pehle chaar terms
$4$	$5$	$4, 9, 14, 19$
$4$	$- 5$	$4, - 1, - 6, - 11$
$- 3$	$0.5$	$- 3, - 2.5, - 2, - 1.5$
$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}$
$7$	$0$	$7, 7, 7, 7$

Aksar kaam ulta hota hai: tumhe ek list di jaati hai aur tay karna hota hai ki woh AP hai ya nahi. Har difference "baad wala term minus pehle wala term" nikaalo. 7, 4, 1, −2, … ke liye hume milta hai $4 - 7 = - 3,$ $1 - 4 = - 3$ aur $- 2 - 1 = - 3.$ Sab barabar hain, isliye yeh AP hai jismein $a = 7$ aur $d = - 3.$ 1, 4, 9, 16, … ke differences 3, 5 aur 7 hain, isliye yeh AP nahi hai. Fig. 5.2 dikhata hai ki barabar aur unequal jumps kaise dikhte hain.

1/4Barabar jumps upar: 2, 6, 10, 14, 18 ek AP hai jismein $d = 4.$

Fig. 5.2AP mein bar tops har baar same jump se chadhte, girte ya same level par rehte hain.

Har step mein paanch terms ke liye paanch bars hain, aur neighbouring bar tops ke beech ka jump neeche likha hai. Step 1: bars 2, 6, 10, 14, 18 plus 4 ke barabar jumps se upar jaate hain, aur ek dashed seedhi line saare tops se guzarti hai. Step 2: bars 18, 14, 10, 6, 2 minus 4 ke barabar jumps se neeche aate hain, phir se ek seedhi line par. Step 3: paanchon bars 6 ke barabar, jumps 0, ek flat line par. Step 4: bars 1, 2, 4, 7, 11, jumps plus 1, plus 2, plus 3, plus 4, dashed bane hain; tops upar ki taraf mudte hain, isliye list par likha hai ki woh AP nahi hai.

Example 1NCERT Example 1 · p. 54

Fractions se a aur d padhna

Yeh rahi fractions se bani ek AP: $- \frac{5}{6}, - \frac{1}{3}, \frac{1}{6}, \frac{2}{3}, \dots$ Iske $a$ aur $d$ kya hain?

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

Example 2NCERT Example 2 · p. 54

AP hai ya nahi?

Har list ko test karo. Agar woh AP hai, to use do terms aur aage badhao. (i) 9, 16, 23, 30, … (ii) 2, −3, −8, −13, … (iii) $\sqrt{5}, \sqrt{20}, \sqrt{45}, \sqrt{80}, \dots$ (iv) 12, 9, 6, 4, …

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

Quick check

4 mein se 1

AP 3.6, 2.9, 2.2, 1.5, … ka common difference kya hai?

Answer dekho

Answer -0.7

2.9 - 3.6 = - 0.7,

aur baaki har pair bhi

- 0.7

deta hai. Terms ghat rahe hain, isliye

d

negative hai.

Practice · 5 questions

Practice 5.1: answers ke saath

1
Short answerEasyNCERT Exercise 5.1, Q1
Tay karo ki har situation se AP banti hai ya nahi, aur samjhao.
1. Ek auto-rickshaw pehle km ke ₹30 aur uske baad har km ke ₹15 leta hai. 1, 2, 3, … km ke baad ka fare dekho.
  Solution dikhao
  Fares: 30, 45, 60, 75, … Har km ₹15 jodta hai, isliye yeh AP hai jismein $d = 15.$
2. Har ghante ek pump tank mein jitna bhi paani hai uska ek-tihaai nikaal deta hai. Har ghante ke baad bacha hua paani dekho.
  Solution dikhao
  Agar tank mein shuru mein $V$ litres hain, to amounts hain $V, \frac{2}{3} V, \frac{4}{9} V, \dots$ Drops $\frac{1}{3} V,$ $\frac{2}{9} V, \dots$ barabar nahi hain (har amount $\frac{2}{3}$ se multiply hota hai), isliye yeh AP nahi hai.
3. Ek stadium stand ki front row mein 40 seats hain aur uske peeche har row mein 4 zyada. Har row ki seats dekho.
  Solution dikhao
  Seats: 40, 44, 48, … Har row 4 jodti hai, isliye yeh AP hai jismein $d = 4.$
4. ₹20,000 ek bank account mein pade hain jis par saal ka 6% compound interest milta hai. Har saal ke end mein balance dekho.
  Solution dikhao
  Amounts: 20000, 21200, 22472, … Increases 1200 aur 1272 barabar nahi hain, kyunki interest par bhi interest milta hai. Yeh AP nahi hai.
2
Short answerEasyNCERT Exercise 5.1, Q2
Har AP ke chaar terms likho: diye gaye $a$ se shuru karo aur har baar diya gaya $d$ jodo.
1. $a = - 6,$ $d = 4$
  Solution dikhao
  $- 6, - 2, 2, 6$
2. $a = 15,$ $d = - 2.5$
  Solution dikhao
  $15, 12.5, 10, 7.5$
3. $a = \frac{1}{2},$ $d = - \frac{1}{3}$
  Solution dikhao
  Sixths mein: $\frac{3}{6}, \frac{1}{6}, - \frac{1}{6}, - \frac{3}{6},$ yaani $\frac{1}{2}, \frac{1}{6}, - \frac{1}{6}, - \frac{1}{2} .$
4. $a = 0,$ $d = - 1.75$
  Solution dikhao
  $0, - 1.75, - 3.5, - 5.25$
3
Short answerEasyNCERT Exercise 5.1, Q3
Har AP ke $a$ aur $d$ padho.
1. 12, 5, −2, −9, … : $a =$ , $d =$
  Answer dekho
  Answer (1) 12, (2) -7
  $d = 5 - 12 = - 7.$
2. −8, −2, 4, 10, … : $a =$ , $d =$
  Answer dekho
  Answer (1) -8, (2) 6
  $d = - 2 - (- 8) = 6.$
3. $\frac{2}{5}, \frac{7}{5}, \frac{12}{5}, \frac{17}{5}, \dots$ : $d =$
  Answer dekho
  Answer 1
  $d = \frac{7}{5} - \frac{2}{5} = \frac{5}{5} = 1,$ aur $a = \frac{2}{5} .$
4. 1.3, 2.1, 2.9, 3.7, … : $d =$
  Answer dekho
  Answer 0.8
  $d = 2.1 - 1.3 = 0.8,$ aur $a = 1.3.$
4
Short answerMediumNCERT Exercise 5.1, Q4
In lists ko APs aur non-APs mein baanto. Har AP ka $d$ batao aur use teen terms aur aage badhao.
1. 3, 6, 12, 24, …
  Solution dikhao
  Differences 3, 6, 12 barabar nahi hain (har term double hota hai). AP nahi hai.
2. $\frac{9}{4}, 2, \frac{7}{4}, \frac{3}{2}, \dots$
  Solution dikhao
  Har difference hai $- \frac{1}{4} .$ AP jismein $d = - \frac{1}{4};$ agle terms $\frac{5}{4}, 1, \frac{3}{4} .$
3. −0.5, −3, −5.5, −8, …
  Solution dikhao
  Har difference hai $- 2.5.$ AP jismein $d = - 2.5;$ agle terms $- 10.5, - 13, - 15.5.$
4. 1, 1.1, 1.21, 1.331, …
  Solution dikhao
  Differences 0.1, 0.11, 0.121 barabar nahi hain (har term pichhle ka 1.1 guna hai). AP nahi hai.
5
Short answerMediumNCERT Exercise 5.1, Q4
In lists ko APs aur non-APs mein baanto. Har AP ka $d$ batao aur use teen terms aur aage badhao.
1. $2 - \sqrt{3}, 2, 2 + \sqrt{3}, 2 + 2 \sqrt{3}, \dots$
  Solution dikhao
  Har difference hai $\sqrt{3} .$ AP jismein $d = \sqrt{3};$ agle terms $2 + 3 \sqrt{3},$ $2 + 4 \sqrt{3},$ $2 + 5 \sqrt{3} .$
2. $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots$
  Solution dikhao
  Differences hain $- \frac{1}{6},$ $- \frac{1}{12},$ $- \frac{1}{20}$ : barabar nahi. AP nahi hai.
3. $7^{2}, 13^{2}, 17^{2}, 409, \dots$
  Solution dikhao
  Terms hain 49, 169, 289, 409. Har difference 120 hai, isliye yeh AP hai jismein $d = 120;$ agle terms 529, 649, 769.
4. $p, 3 p, 5 p, 7 p, \dots$
  Solution dikhao
  Har difference hai $2 p .$ AP jismein $d = 2 p;$ agle terms $9 p, 11 p, 13 p .$

5.3 AP ka $n$ th Term

Tumhara school ek auditorium bana raha hai. Row 1 mein 18 seats hongi, aur har row mein apne aage wali row se 4 seats zyada hongi. Row 15 mein kitni seats hongi? Tum row dar row baar-baar 4 jod sakte ho, lekin ek tez tareeka bhi hai.

Har row ko 18 plus kuch 4s ki tarah likho. Row 2 mein $18 + 1 \times 4$ seats hain, row 3 mein $18 + 2 \times 4$ aur row 4 mein $18 + 3 \times 4.$ 4s ki ginti hamesha row number se ek kam hoti hai, kyunki 4 tabhi judta hai jab tum ek row peeche jaate ho. To row 15 mein $18 + 14 \times 4 = 74$ seats hain, aur row 30 mein hongi $18 + 29 \times 4 = 134.$

3/4Row 5 chaar jumps aage hai: $18 + 4 \times 4 = 34.$

Fig. 5.3

n

th term tak pahunchne ke liye tum first term se

d

n - 1

jumps lagaate ho.

Chhe bars rows 1 se 6 ke liye hain, jinmein 18, 22, 26, 30, 34 aur 38 seats hain. Step 1 pehle bar, 18, ko highlight karta hai, jo first term a hai. Step 2 row 1 se row 2 tak plus 4 ka ek jump dikhata hai, yaani 18 plus 4. Step 3 row 1 se row 5 tak ke jumps dikhata hai aur unhe 1, 2, 3, 4 number deta hai: row 5, row 1 se chaar jumps baad hai, isliye usmein 18 plus 4 times 4, yaani 34 seats hain. Step 4 rule batata hai: row n tak pahunchne ke liye tum n minus 1 jumps lagaate ho, isliye nth term hai a plus (n minus 1) times d.

DerivationAP ka

n

th term

Maan lo ek AP ka first term hai $a$ aur common difference $d .$ Har ek step aage badhne par ek $d$ judta hai:

\begin{aligned} a_{1} & = a \\ a_{2} & = a + d \\ a_{3} & = a + 2 d \\ a_{4} & = a + 3 d \end{aligned}

Har line mein $d$ ki ginti position se ek kam hai. First term se $n$ th term tak pahunchne ke liye tum $n - 1$ steps lete ho, isliye

AP ka $n$th term

a_{n} = a + (n - 1) d

(5.4)

jahan

$a$: first term
$d$: common difference
$n$: term ki position (1, 2, 3, …)

Kyunki isse har term milta hai, $a_{n}$ ko general term kehte hain. Finite AP mein last term ka apna letter hota hai, $l$ : $m$ terms hon to $l = a_{m} .$ $a_{n},$ $a,$ $d$ aur $n$ mein se koi bhi ek unknown ho sakta hai, aur neeche ke examples baari-baari se har ek ke liye solve karte hain.

InteractiveAP explorer

Arithmetic progressions ke liye ek interactive explorer. Tum first term a, common difference d aur number of terms n ko minus aur plus buttons se ya type karke set karte ho, ya koi preset chunte ho, jaise auditorium seats 18, plus 4. Terms view mein ek bar chart first n terms dikhata hai, positive terms ke bars line ke upar aur negative terms ke neeche, aur last bar outlined hota hai; Play button bars ko ek-ek karke jodta hai. Ek panel tumhare numbers ke saath nth term ko a plus (n minus 1) times d ki tarah nikaalta hai. Sum view mein ek button har bar par ek doosra, dashed orange tukda jodta hai, jo ulti likhi list se liya gaya hai, taaki har column same height tak pahunche, yaani first term plus last term; panel S n ko n halves times (a plus l) ki tarah nikaalta hai. Ek table har term ko running total ke saath dikhata hai, aur ek note batata hai jab koi pehle ka running total last wale ke barabar ho.

Fig. 5.4Ek first term, ek common difference aur number of terms chuno;

n

th term aur sum padho.

Example 3NCERT Example 3 · p. 58

Door ka term nikaalna

8, 14, 20, … ka 16th term kya hai?

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

Example 4NCERT Example 4 · p. 58

Yeh kaun sa term hai?

AP 64, 59, 54, … mein −11 kahan aata hai? Kya 0 kabhi aata hai?

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

Poori AP nikaalne ke liye uske do terms kaafi hain. Unke beech ka gap jumps ki ek poori ginti hota hai, jaisa Fig. 5.5 dikhata hai.

3/3 $a_{4}$ se teen jumps peeche: $a = 11 - 9 = 2.$

Fig. 5.54th term se 9th term tak

9 - 4 = 5

jumps hain.

AP ki positions 1 se 9 ek row mein boxes ki tarah bani hain. Step 1 sirf box 4 mein 11 aur box 9 mein 26 bharta hai; baaki question marks hain. Step 2 box 4 se box 9 tak paanch barabar jump arrows banata hai aur likhta hai ki 5 jumps 26 minus 11, yaani 15, cover karte hain, isliye har jump d 3 hai. Step 3 box 4 se box 1 tak teen jumps peeche jaata hai, isliye first term 11 minus 9, yaani 2 hai, aur har box bhar deta hai: 2, 5, 8, 11, 14, 17, 20, 23, 26.

Example 5NCERT Example 5 · p. 58

Do terms se AP

Ek AP ka 4th term 11 hai aur 9th term 26 hai. AP nikaalo.

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

Quick check

2 mein se 1

AP 4, 9, 14, … ka 20th term nikaalo.

Answer dekho

Answer 99

a_{20} = 4 + 19 \times 5 = 99.

19 ki jagah 20 jumps lene se galat answer 104 aata hai.

Example 6NCERT Example 6 · p. 59

Kya yeh term hai?

Kya 254 list 8, 15, 22, 29, … ka hissa hai?

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

Example 7NCERT Example 7 · p. 59

Formula se ginti

Kitne two-digit numbers 7 se divide karne par remainder 1 chhodte hain?

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

Example 8NCERT Example 8 · p. 59

End se ginti

AP 3, 7, 11, …, 119 mein 119 se peeche ki taraf ginne par 8th number kaun sa hai?

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

2/3End se 8th term ki position hai $30 - 8 + 1 = 23.$

Fig. 5.6

n

terms wali AP ka end se

k

th term uska

(n - k + 1)

th term hota hai.NCERT Example 8 · p. 59

Step 1 AP ko boxes ki ek row ki tarah dikhata hai, neeche positions 1, 2, 3 aur aise hi 30 tak likhi hain; pehle boxes mein 3, 7, 11 hain aur last mein 119. Step 2 boxes ke upar right se chalne wali doosri ginti jodta hai: end se 1st, 2nd, 3rd, aur end se 8th position 23 par aata hai, jis par ek ring laga hai. Step 3 right se left ek arrow jodta hai: ulta padhne par AP 119 se shuru hoti hai aur 4 se ghatti hai, aur ring wale box mein 91 hai.

Example 9NCERT Example 9 · p. 60

Simple interest se AP banti hai

Neha ₹6000 saal ke 7% simple interest par jama karti hai. Dikhao ki saal 1, 2, 3, … ke end mein uske account ke amounts ek AP banate hain, aur 15 saal baad ka amount nikaalo.

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

Example 10NCERT Example 10 · p. 61

Diyon ki rows

Diwali par diye rows mein rakhe jaate hain: pehli row mein 41, doosri mein 37, teesri mein 33, aur aise hi aage, aur last row mein 9. Rows ki ginti nikaalo.

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

Quick check

3 mein se 1

AP 100, 94, 88, … mein 28 kahan aata hai?

11th12th13th14th

Answer dekho

Answer (C) 13th

6 (n - 1) = 72,

isliye

n - 1 = 12

aur

n = 13.

Answer 12th tab aata hai jab tum 1 wapas jodna bhool jaate ho.

5.4 AP ke First $n$ Terms ka Sum

Wapas auditorium par chalte hain. Maan lo usmein 25 rows hain. Kul kitni seats hain? Rows mein 18, 22, 26, … seats hain, isliye hume chahiye $18 + 22 + 26 + \dots$ 25 terms tak. Inhe ek-ek karke jodna slow hai aur galti hona aasaan hai. Hume ek formula chahiye.

Pairing wali trick har AP ke liye kaam karti hai, sirf 1, 2, 3, … ke liye nahi. Dekho kyun.

DerivationFirst

n

terms ka sum

Koi bhi AP lo. Maan lo uska first term hai $a,$ common difference $d,$ aur $S_{n}$ ka matlab hai $a_{1} + a_{2} + \dots + a_{n} .$ $n$ th term ko kaho $l,$ yaani $l = a + (n - 1) d .$ Total ko pehle aage se likho, phir peeche se (dots beech ke terms ki jagah hain):

\begin{aligned} S_{n} & = a + \dots + l \\ S_{n} & = l + \dots + a \end{aligned}

Dono lines ko column dar column jodo. Har column mein ek $+ d$ ek $- d$ se milta hai: pehla column hai $a + l,$ doosra $(a + d) + (l - d) = a + l,$ teesra $(a + 2 d) + (l - 2 d) = a + l,$ aur aise hi aage. Kul $n$ columns hain, sab $a + l$ ke barabar:

2 S_{n} = n (a + l)

2 se divide karke (5.5) milta hai; phir $l = a + (n - 1) d$ rakhne par (5.6) milta hai.

Sum jab last term pata ho

S_{n} = \frac{n}{2} (a + l)

(5.5)

First $n$ terms ka sum

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

(5.6)

jahan

$n$: jode gaye terms ki ginti
$a$: first term
$d$: common difference
$l$: last ( $n$ th) term

4/4 $2 S_{5} = 5 \times 16 = 80,$ isliye $S_{5} = 40.$

Fig. 5.7Sum ki do copies, ek ulti,

n

columns chaudi aur

a + l

oonchi ek rectangle bhar deti hain.NCERT p. 64

Step 1 mein 2, 5, 8, 11 aur 14 height ke paanch blue bars hain, yaani jodne wali AP. Step 2 wahi paanch values ulte order mein, 14, 11, 8, 5, 2, dashed orange bars ki tarah dikhata hai jo upar ki ek line se latke hain. Step 3 orange bars ko neeche blue bars par khiskaata hai, to columns ban jaate hain 2 plus 14, 5 plus 11, 8 plus 8, 11 plus 5 aur 14 plus 2, har ek 16 ooncha. Step 4 rectangle ki outline banata hai, 5 columns by 16, jismein 80 units hain: sum ka do guna, isliye sum 40 hai, jo 5 over 2 times (2 plus 14) hai.

Jab first aur last terms pata hon to (5.5) use karo, aur (5.6) tab jab pata hon $a$ aur $d .$ Auditorium ke liye $a = 18,$ $d = 4$ aur $n = 25,$ isliye last row mein $18 + 24 \times 4 = 114$ seats hain aur

\begin{aligned} S_{25} & = \frac{25}{2} (18 + 114) \\ = 25 \times 66 = 1650 \end{aligned}

Hall mein 1650 log baith sakte hain. Jaise $a_{n},$ waise hi $S_{n},$ $a,$ $d$ aur $n$ mein se koi bhi ek unknown ho sakta hai, jaisa neeche ke examples dikhate hain.

Ek aur kaam ki baat. First $n$ terms ka sum, first $n - 1$ terms ke sum plus ek aur term, yaani $n$ th term, ke barabar hai. Isliye sums ke formula se tum koi bhi term nikaal sakte ho (aur $a_{1} = S_{1}$ ):

Sums se $n$th term

a_{n} = S_{n} - S_{n - 1}

(5.7)

Quick check

2 mein se 1

Reverse-and-add argument ke steps ko sahi order mein rakho.

Har column $a + l$ hai
Dono lines ko column dar column jodo
$n (a + l)$ ko 2 se divide karo
$S_{n}$ aage se likho
$S_{n}$ peeche se likho

Answer dekho

Answer

$S_{n}$ aage se likho
$S_{n}$ peeche se likho
Dono lines ko column dar column jodo
Har column $a + l$ hai
$n (a + l)$ ko 2 se divide karo

Aage aur peeche se likhne par milta hai

2 S_{n} = n (a + l),

kyunki

n

columns mein se har ek ka total hai

a + l .

Aadha karne par milta hai

S_{n} = \frac{n}{2} (a + l) .

Example 11NCERT Example 11 · p. 65

Negative common difference wala sum

15, 11, 7, … ke first 18 terms jodo.

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

Example 12NCERT Example 12 · p. 65

Sum se term tak

Ek AP 5 se shuru hoti hai, aur uske first 12 terms ka sum 456 hai. Uska 15th term kya hai?

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

Example 13NCERT Example 13 · p. 66

Number of terms ke do answers

First term se shuru karke, 32, 28, 24, … ke kitne terms ka sum 132 hota hai?

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

2/3 $8 + 4 + 0 + (- 4) + (- 8) = 0$ : yeh terms cancel ho jaate hain.

Fig. 5.8Terms 7 se 11 aapas mein cancel ho jaate hain, isliye first 6 aur first 11 terms ka sum same hai.NCERT Example 13 · p. 66

Ek horizontal line par terms 32, 28, 24, 20, 16, 12, 8, 4, 0, −4, −8 ke gyaarah bars khade hain; last do line ke neeche latke hain. Step 1 pehle chhe bars ko shade karta hai aur unke total par 132 likhta hai. Step 2 bars 7 se 11, yaani terms 8, 4, 0, −4 aur −8, ko highlight karta hai aur 8 ko −8 se aur 4 ko −4 se jodta hai, yeh dikhane ke liye ki woh cancel hote hain, aur beech mein 0 hai. Step 3 saare gyaarah bars ko shade karta hai aur total par phir se 132 likhta hai: S6, S11 ke barabar hai.

Example 14NCERT Example 14 · p. 66

Counting numbers ke sums

Jodo (i) 1 se 250 tak ke saare whole numbers aur (ii) saare whole numbers 1 se lekar $n .$

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

Example 15NCERT Example 15 · p. 67

Terms ke formula se sum

Ek list is rule se di gayi hai: $a_{n} = 7 - 3 n .$ Dikhao ki yeh AP hai aur uske first 20 terms jodo.

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

Example 16NCERT Example 16 · p. 67

Badhti hui school library

Ek school library har saal jitni nayi books khareedti hai, woh har saal same amount se badhti hain. Usne apne 2nd saal mein 340 nayi books khareedi aur 6th saal mein 460. Nikaalo (i) 1st saal mein khareedi books, (ii) 12th saal mein khareedi books aur (iii) pehle 8 saalon mein khareedi kul books.

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

Example

Badhte instalments mein loan chukaana

Kiran ek self-help group se ₹12,600 interest-free udhaar leti hai. Woh pehle mahine ₹500 chukaati hai aur har mahine pichhle mahine se ₹100 zyada deti hai. Loan chukaane mein use kitne mahine lagte hain, aur uski last instalment kitni hai?

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

Quick check

3 mein se 1

Ek AP mein 19 terms hain. Uska first term 6 hai aur last term 60 hai. Uska sum hai

570

627

660

1254

Answer dekho

Answer (B) $627$

S_{19} = \frac{19}{2} \times 66 = 19 \times 33 = 627.

1254 mein aadha karna bhool gaye; 570 aur 660 sirf ek end ka 19 guna lete hain.

1/3Practice 5.3 Q19: radii 1, 2, 3, … cm hote jaate hain.

Fig. 5.9Practice 5.3 Q19 aur Q21 aur Practice 5.4 Q5 ke liye pictures.NCERT Fig. 5.4–5.8

Step 1 quarter circles se bana ek spiral dikhata hai. Pehle ka radius 1 cm hai, aur har agla quarter circle usi taraf mudta hai aur uska radius 1 cm zyada hota hai; inpar 1, 2, 3 aur 4 likha hai. Step 2 ek fielding drill dikhata hai: left mein stumps ka ek set aur right mein ek seedhi line mein cricket balls, pehli stumps se 6 m door aur baaki 4 m ke gap par, aur dashed arrows ek ball tak daud kar wapas aane ka ek round dikhate hain. Step 3 river ke paas ghat ki seedhiyon ka side view dikhata hai, baarah mein se saat bani hain: har step 0.5 m gehri hai aur neeche wali se 0.2 m oonchi, aur har step zameen tak ek solid block hai.

Practice · 21 questions

Practice 5.3: answers ke saath

1
Short answerEasyNCERT Exercise 5.3, Q1
Har AP ke diye gaye number of terms jodo.
1. 3, 9, 15, … 12 terms tak
  Solution dikhao
  $S_{12} = 6 (6 + 11 \times 6) = 6 \times 72 = 432.$
2. −40, −35, −30, … 15 terms tak
  Solution dikhao
  $S_{15} = \frac{15}{2} (- 80 + 14 \times 5) = \frac{15}{2} \times (- 10) = - 75.$
3. 0.4, 1.3, 2.2, … 50 terms tak
  Solution dikhao
  $d = 0.9.$ $S_{50} = 25 (0.8 + 49 \times 0.9) = 25 \times 44.9 = 1122.5.$
4. $\frac{3}{4}, \frac{1}{2}, \frac{1}{4}, \dots$ 9 terms tak
  Solution dikhao
  $d = - \frac{1}{4} .$ $S_{9} = \frac{9}{2} (\frac{3}{2} - 2) = \frac{9}{2} \times (- \frac{1}{2}) = - \frac{9}{4} .$
2
Short answerMediumNCERT Exercise 5.3, Q2
Har sum nikaalo.
1. $4 + 6 \frac{1}{2} + 9 + \dots + 54$
  Solution dikhao
  $d = 2 \frac{1}{2} .$ $54 = 4 + 2.5 (n - 1)$ se milta hai $n = 21.$ $S = \frac{21}{2} \times 58 = 609.$
2. $50 + 47 + 44 + \dots + 2$
  Solution dikhao
  $2 = 50 - 3 (n - 1)$ se milta hai $n = 17.$ $S = \frac{17}{2} \times 52 = 442.$
3. $- 2 + (- 6) + (- 10) + \dots + (- 118)$
  Solution dikhao
  $- 118 = - 2 - 4 (n - 1)$ se milta hai $n = 30.$ $S = 15 \times (- 120) = - 1800.$
3
Short answerMediumNCERT Exercise 5.3, Q3
Har part mein AP ke baare mein kuch facts diye hain. Jo do quantities poochhi gayi hain, unhe nikaalo.
1. $a = 4,$ $d = 5,$ $a_{n} = 89$ : $n$ aur $S_{n}$ nikaalo
  Solution dikhao
  $89 = 4 + 5 (n - 1)$ se milta hai $n = 18.$ $S_{18} = 9 (4 + 89) = 837.$
2. $a = 9,$ $a_{15} = 51$ : $d$ aur $S_{15}$ nikaalo
  Solution dikhao
  $14 d = 42,$ isliye $d = 3.$ $S_{15} = \frac{15}{2} (9 + 51) = 450.$
3. $a_{10} = 47,$ $d = 4$ : $a$ aur $S_{10}$ nikaalo
  Solution dikhao
  $a = 47 - 9 \times 4 = 11.$ $S_{10} = 5 (11 + 47) = 290.$
4. $a_{4} = 17,$ $S_{12} = 264$ : $d$ aur $a_{12}$ nikaalo
  Solution dikhao
  $a + 3 d = 17$ aur $6 (2 a + 11 d) = 264,$ isliye $2 a + 11 d = 44.$ Phir $2 (17 - 3 d) + 11 d = 44$ se milta hai $5 d = 10,$ $d = 2,$ $a = 11$ aur $a_{12} = 11 + 22 = 33.$
4
Short answerHardNCERT Exercise 5.3, Q3
Har part mein AP ke baare mein kuch facts diye hain. Jo do quantities poochhi gayi hain, unhe nikaalo.
1. $d = 4,$ $S_{8} = 136$ : $a$ aur $a_{8}$ nikaalo
  Solution dikhao
  $4 (2 a + 28) = 136$ se milta hai $a = 3.$ $a_{8} = 3 + 28 = 31.$
2. $a = 3,$ $d = 6,$ $S_{n} = 192$ : $n$ aur $a_{n}$ nikaalo
  Solution dikhao
  $\frac{n}{2} [6 + 6 (n - 1)] = 3 n^{2} = 192,$ isliye $n^{2} = 64$ aur $n = 8.$ $a_{8} = 3 + 42 = 45.$
3. $a = 6,$ $a_{n} = 72,$ $S_{n} = 468$ : $n$ aur $d$ nikaalo
  Solution dikhao
  $\frac{n}{2} (6 + 72) = 468$ se milta hai $n = 12.$ $72 = 6 + 11 d$ se milta hai $d = 6.$
4. $a_{n} = 3,$ $d = 2,$ $S_{n} = - 12$ : $n$ aur $a$ nikaalo
  Solution dikhao
  $a = 3 - 2 (n - 1) = 5 - 2 n .$ Phir $S_{n} = \frac{n}{2} (5 - 2 n + 3) = n (4 - n) = - 12,$ isliye $n^{2} - 4 n - 12 = 0,$ $(n - 6) (n + 2) = 0$ aur $n = 6.$ To $a = - 7.$
Short answerMediumNCERT Exercise 5.3, Q4
7, 11, 15, … ke terms ko shuru se order mein jodte hue, kitne terms ke baad total theek 900 hota hai?
Solution dikhao
$900 = \frac{n}{2} [14 + 4 (n - 1)] = n (2 n + 5),$ isliye $2 n^{2} + 5 n - 900 = 0.$
Discriminant hai $25 + 7200 = 7225 = 85^{2},$ isliye $n = \frac{- 5 + 85}{4} = 20$ (doosra root negative hai).
Short answerMediumNCERT Exercise 5.3, Q5
Ek AP 8 se shuru hoti hai, 68 par khatam hoti hai aur uske terms ka sum 608 hai. Usmein kitne terms hain, aur $d$ kya hai?
Solution dikhao
$608 = \frac{n}{2} (8 + 68) = 38 n,$ isliye $n = 16.$
$68 = 8 + 15 d$ se milta hai $d = 4.$
Short answerMediumNCERT Exercise 5.3, Q6
Ek AP 11 se 263 tak jaati hai, aur har step hai $d = 6.$ Pata karo usmein kitne terms hain aur unka total kya hai.
Solution dikhao
$263 = 11 + 6 (n - 1)$ se milta hai $n - 1 = 42,$ isliye $n = 43.$
$S_{43} = \frac{43}{2} (11 + 263) = 43 \times 137 = 5891.$
Short answerEasyNCERT Exercise 5.3, Q7
Ek AP mein $d = 5$ hai aur uska 25th term 130 hai. Uske first 25 terms ka sum kya hai?
Solution dikhao
$a = 130 - 24 \times 5 = 10.$ $S_{25} = \frac{25}{2} (10 + 130) = 1750.$
Short answerMediumNCERT Exercise 5.3, Q8
Ek AP ke 2nd aur 3rd terms 11 aur 15 hain. Nikaalo $S_{40} .$
Solution dikhao
$d = 4$ aur $a = 11 - 4 = 7.$ $S_{40} = 20 (14 + 39 \times 4) = 20 \times 170 = 3400.$
Short answerHardNCERT Exercise 5.3, Q9
Ek AP mein $S_{6} = 48$ aur $S_{12} = 168.$ Iske liye formula nikaalo: $S_{n} .$
Solution dikhao
$3 (2 a + 5 d) = 48$ se milta hai $2 a + 5 d = 16;$ $6 (2 a + 11 d) = 168$ se milta hai $2 a + 11 d = 28.$
Ghatane par $6 d = 12,$ isliye $d = 2$ aur $a = 3.$
$S_{n} = \frac{n}{2} [6 + 2 (n - 1)] = n (n + 2) = n^{2} + 2 n .$ Check: $S_{6} = 48$ aur $S_{12} = 168.$
11
Short answerMediumNCERT Exercise 5.3, Q10
Neeche ka har rule ek list $a_{1}, a_{2}, a_{3}, \dots$ define karta hai. Prove karo ki list AP hai aur uske first 12 terms jodo.
1. $a_{n} = 5 + 3 n$
  Solution dikhao
  $a_{n + 1} - a_{n} = 3$ hai, chahe jo bhi ho $n,$ isliye yeh AP hai jismein $d = 3.$ $a_{1} = 8,$ $a_{12} = 41,$ isliye $S_{12} = 6 (8 + 41) = 294.$
2. $a_{n} = 11 - 4 n$
  Solution dikhao
  $a_{n + 1} - a_{n} = - 4$ hai, chahe jo bhi ho $n,$ isliye yeh AP hai jismein $d = - 4.$ $a_{1} = 7,$ $a_{12} = - 37,$ isliye $S_{12} = 6 (7 - 37) = - 180.$
12
Short answerHardNCERT Exercise 5.3, Q11
Ek AP ke liye $S_{n} = 3 n^{2} + 2 n$ hai, chahe jo bhi ho $n .$ Nikaalo:
1. first term
  Solution dikhao
  $a_{1} = S_{1} = 3 + 2 = 5.$
2. second term
  Solution dikhao
  $S_{2} = 12 + 4 = 16,$ isliye $a_{2} = 16 - 5 = 11.$
3. $n$ th term
  Solution dikhao
  $a_{n} = (3 n^{2} + 2 n) - [3 (n - 1)^{2} + 2 (n - 1)] = 6 n - 1.$ (Check: $a_{1} = 5,$ $a_{2} = 11.$ )
4. 10th term
  Solution dikhao
  $a_{10} = 6 \times 10 - 1 = 59.$
Short answerEasyNCERT Exercise 5.3, Q12
7 ke pehle 30 positive multiples jodo.
Solution dikhao
$S_{30} = \frac{30}{2} (7 + 210) = 15 \times 217 = 3255.$
Short answerEasyNCERT Exercise 5.3, Q13
9 ke pehle 20 multiples (yaani 9, 18, 27, …) ka sum kya hai?
Solution dikhao
$S_{20} = 10 (9 + 180) = 1890.$
Short answerMediumNCERT Exercise 5.3, Q14
101 se 199 tak ke saare odd numbers jodo.
Solution dikhao
Woh hain 101, 103, …, 199. $199 = 101 + 2 (n - 1)$ se milta hai $n = 50.$
$S_{50} = 25 (101 + 199) = 7500.$
Short answerEasyNCERT Exercise 5.3, Q15
Meena ek savings plan shuru karti hai: pehle week ₹100, doosre mein ₹120, teesre mein ₹140, aur aise hi aage. 26 weeks baad usne kitne bachaaye honge?
₹
Solution dikhao
$a = 100,$ $d = 20,$ $n = 26.$ $S_{26} = 13 (200 + 25 \times 20) = 13 \times 700 = 9100.$ Woh ₹9100 bachaati hai.
Short answerMediumNCERT Exercise 5.3, Q16
Ek quiz club ke paas 8 prizes ke liye ₹1600 hain. Har prize apne theek upar wali rank ke prize se ₹30 kam hai. Har prize kitne ka hai?
Solution dikhao
$1600 = 4 [2 a + 7 \times (- 30)]$ se milta hai $2 a - 210 = 400,$ isliye $a = 305.$
Prizes (₹ mein) hain 305, 275, 245, 215, 185, 155, 125 aur 95.
Short answerMediumNCERT Exercise 5.3, Q17
Ek eco-club drive mein Class 6 se Class 12 tak har class ka har section utne saplings lagata hai jitna uska class number hai (Class 6 ka section 6 lagata hai). Har class mein 4 sections hain. Kul kitne saplings lagaaye jaate hain?
Solution dikhao
Har class ka ek section lagata hai $6 + 7 + \dots + 12 = \frac{7}{2} (6 + 12) = 63.$
Har class mein 4 sections ke saath: $4 \times 63 = 252$ saplings.
NumericalMediumNCERT Exercise 5.3, Q18
12 quarter circles se ek spiral banaya gaya hai (Fig. 5.9, step 1). Pehle ka radius 1 cm hai aur har agle ka radius 1 cm zyada hai. Spiral ki kul length nikaalo. (Lo $π = \frac{22}{7} .$ )
cm
Solution dikhao
Lengths hain $\frac{π}{2} \times 1,$ $\frac{π}{2} \times 2,$ …, $\frac{π}{2} \times 12,$ ek AP.
Total $= \frac{π}{2} (1 + 2 + \dots + 12) = \frac{π}{2} \times 78 = 39 π .$
$π = \frac{22}{7}$ lene par: $39 \times \frac{22}{7} = \frac{858}{7} \approx 122.57 c m .$
Short answerHardNCERT Exercise 5.3, Q19
Ek builder 150 bricks ka dher lagata hai: sabse neeche ki row mein 18, uske upar ki row mein 17, agli mein 16, aur aise hi aage. Dher kitni rows ooncha hai, aur sabse upar ki row mein kitni bricks hain?
Solution dikhao
$150 = \frac{n}{2} [36 - (n - 1)] = \frac{n (37 - n)}{2},$ isliye $n^{2} - 37 n + 300 = 0$ aur $(n - 12) (n - 25) = 0.$
25 rows hone par sabse upar ki row mein $18 - 24 = - 6$ bricks hongi, jo impossible hai. Isliye 12 rows hain aur sabse upar ki row mein $18 - 11 = 7$ bricks hain.
NumericalMediumNCERT Exercise 5.3, Q20
Ek fielding drill mein (Fig. 5.9, step 2), das balls stumps se ek seedhi line mein padi hain: pehli 6 m door hai aur baaki aapas mein 4 m ke gap par hain. Ek fielder stumps se shuru karti hai, sabse paas wali ball tak daudti hai, use stumps par wapas laati hai, phir agli laati hai, aur aise hi jab tak dason wapas na aa jaayein. Woh kitna daudti hai?
m
Solution dikhao
Balls 6, 10, 14, …, 42 m door hain, ek AP jismein $a = 6,$ $d = 4,$ aur 10 terms. Unka sum hai $5 (6 + 42) = 240.$
Har trip jaana aur wapas aana hai, isliye woh daudti hai $2 \times 240 = 480 m .$

NCERT ne apni Exercise 5.4 ko optional mark kiya hai: yeh board examination ke liye nahi hai. Neeche ke questions wahi skills practice karate hain, jab tumhe challenge chahiye.

Practice · 5 questions

Practice 5.4 (optional): answers ke saath

Short answerMediumNCERT Exercise 5.4 (optional), Q1
AP 150, 143, 136, … aakhir mein zero se neeche chali jaati hai. Pehla negative term kaun sa hai?
Solution dikhao
$150 - 7 (n - 1) < 0$ se milta hai $n - 1 > 21.4,$ isliye aisa sabse chhota $n$ 23 hai.
Check: $a_{22} = 150 - 147 = 3$ aur $a_{23} = - 4.$
Short answerHardNCERT Exercise 5.4 (optional), Q2
Ek AP mein $a_{4} + a_{8} = 20$ aur $a_{4} \times a_{8} = 64.$ Nikaalo $S_{12} .$
Solution dikhao
$a_{4} + a_{8} = 2 a_{6} = 20,$ isliye $a_{6} = 10.$ Likho $a_{4} = 10 - 2 d$ aur $a_{8} = 10 + 2 d .$
$(10 - 2 d) (10 + 2 d) = 100 - 4 d^{2} = 64,$ isliye $d^{2} = 9$ aur $d = 3$ ya $d = - 3.$
Phir $a = 10 - 5 d$ aur $S_{12} = 6 (2 a + 11 d) = 6 (20 + d) .$ Agar $d = 3,$ to $S_{12} = 138;$ agar $d = - 3,$ to $S_{12} = 102.$ Dono APs diye gaye facts par fit hoti hain.
NumericalMediumNCERT Exercise 5.4 (optional), Q3
Ek wooden wall rack ki shelves 20 cm ke gap par hain. Shelves har baar same amount se chhoti hoti jaati hain, neeche 90 cm se upar 30 cm tak, aur sabse upar aur sabse neeche ki shelves 1.4 m door hain. Shelves ke liye kul kitni length ki lakdi chahiye?
cm
Solution dikhao
Kul $\frac{140}{20} + 1 = 8$ shelves hain.
Total $= \frac{8}{2} (90 + 30) = 480 c m = 4.8 m .$
ProofHardNCERT Exercise 5.4 (optional), Q4
Ek lambi street par houses ke numbers 1 se 288 tak hain. Woh house $x$ nikaalo jiske neeche ke saare numbers ka total, $x$ ke upar ke saare numbers ke total ke barabar ho; yaani total dono taraf barabar hai, aur beech mein hai $x .$
Solution dikhao
Hume chahiye $\frac{(x - 1) x}{2} = \frac{288 \times 289}{2} - \frac{x (x + 1)}{2} .$
2 se multiply karo aur terms ikattha karo: $x^{2} - x + x^{2} + x = 288 \times 289,$ isliye $x^{2} = 144 \times 289 = 41616$ aur $x = 12 \times 17 = 204.$
Check: $1 + \dots + 203 = 20706$ aur $205 + \dots + 288 = 20706.$ Aisi ek puzzle 1914 mein ek magazine mein chhapi thi. Jab P. C. Mahalanobis ne use Srinivasa Ramanujan ko padh kar sunaaya, to unhone ek hi baar mein har street length ke liye uska answer de diya.
NumericalHardNCERT Exercise 5.4 (optional), Q5
River ghat ki seedhiyaan (Fig. 5.9, step 3): 12 steps, har ek 40 m lambi, 0.2 m oonchi uthti hai aur 0.5 m gehri hai. Har step zameen se upar tak concrete ka ek solid block hai. Concrete ka volume nikaalo.
m³
Solution dikhao
Step $k$ hai $0.2 k \times 0.5 \times 40 = 4 k m^{3} .$
Total $= 4 (1 + 2 + \dots + 12) = 4 \times 78 = 312 m^{3} .$

5.5 Summary

3/3 $S_{6} = \frac{6}{2} (4 + 19) = 69.$

Fig. 5.10Poora chapter ek AP par: barabar jumps, koi bhi term, sum.

Ek AP, 4, 7, 10, 13, 16, 19, chhe bars ki tarah bani hai. Step 1 bar tops ke beech plus 3 ke barabar jumps dikhata hai aur first term a = 4 aur common difference d = 3 likhta hai. Step 2 chhathe bar ko highlight karta hai aur dikhata hai ki woh pehle bar se paanch jumps upar hai: a6 = 4 + 5 times 3 = 19. Step 3 bars ke upar ek ulti copy jodta hai, taaki har column 4 + 19 = 23 tak pahunche, aur chhe terms ka sum 6 times 23 divided by 2, yaani 69 hai.

Recap

Summary aur key points

AP mein tum har term se agle term tak same number $d$ (common difference) jod kar pahunchte ho, jo positive, negative ya zero ho sakta hai (Section 5.2).
Kisi list ko test karne ke liye har gap "baad wala minus pehle wala", $a_{k + 1} - a_{k}$ , nikaalo; ek bhi alag gap ho to woh AP nahi hai ((5.1)).
General form: $a, a + d, a + 2 d, \dots$ Finite AP ka ek last term $l$ hota hai; infinite AP ka nahi hota ((5.2)).
Agar $a, b, c$ AP mein hain, to $b = \frac{a + c}{2},$ yaani $a$ aur $c$ ka arithmetic mean ((5.3)).
$n$ th term hai $a_{n} = a + (n - 1) d$ : first term ke baad $n - 1$ jumps ((5.4)).
Koi number tabhi term hai jab $a_{n} = x$ solve karne par counting number $n$ mile (Example 4).
Do terms se AP tay ho jaati hai: unke beech ka gap jumps ki poori ginti hota hai (Example 5).
$n$ terms wali AP ka end se $k$ th term uska $(n - k + 1)$ th term hai, ya $l - (k - 1) d$ (Example 8).
Sum ko aage aur peeche se likhne par milta hai $S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} [2 a + (n - 1) d]$ (sum ka derivation).
$a_{n} = S_{n} - S_{n - 1},$ aur $1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ ((5.7), Example 14).
"Kitne terms" wala sawaal $n$ mein ek quadratic tak le jaata hai; har woh root rakho jo counting number ho aur story mein fit ho (Example 13).

Board-exam practice

Yeh questions CBSE board paper ke formats follow karte hain: multiple choice, assertion–reason, short aur long answers, aur ek case study.

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 5.1, Q1	Practice 5.1 · Q1
NCERT Exercise 5.1, Q2	Practice 5.1 · Q2
NCERT Exercise 5.1, Q3	Practice 5.1 · Q3
NCERT Exercise 5.1, Q4	Practice 5.1 · Q4
NCERT Exercise 5.1, Q4	Practice 5.1 · Q5
NCERT Exercise 5.2, Q1	Practice 5.2 · Q1
NCERT Exercise 5.2, Q2	Practice 5.2 · Q2
NCERT Exercise 5.2, Q3	Practice 5.2 · Q3
NCERT Exercise 5.2, Q4	Practice 5.2 · Q4
NCERT Exercise 5.2, Q5	Practice 5.2 · Q5
NCERT Exercise 5.2, Q6	Practice 5.2 · Q6
NCERT Exercise 5.2, Q7	Practice 5.2 · Q7
NCERT Exercise 5.2, Q8	Practice 5.2 · Q8
NCERT Exercise 5.2, Q9	Practice 5.2 · Q9
NCERT Exercise 5.2, Q10	Practice 5.2 · Q10
NCERT Exercise 5.2, Q11	Practice 5.2 · Q11
NCERT Exercise 5.2, Q12	Practice 5.2 · Q12
NCERT Exercise 5.2, Q13	Practice 5.2 · Q13
NCERT Exercise 5.2, Q14	Practice 5.2 · Q14
NCERT Exercise 5.2, Q15	Practice 5.2 · Q15
NCERT Exercise 5.2, Q16	Practice 5.2 · Q16
NCERT Exercise 5.2, Q17	Practice 5.2 · Q17
NCERT Exercise 5.2, Q18	Practice 5.2 · Q18
NCERT Exercise 5.2, Q19	Practice 5.2 · Q19
NCERT Exercise 5.2, Q20	Practice 5.2 · Q20
NCERT Exercise 5.3, Q1	Practice 5.3 · Q1
NCERT Exercise 5.3, Q2	Practice 5.3 · Q2
NCERT Exercise 5.3, Q3	Practice 5.3 · Q3
NCERT Exercise 5.3, Q3	Practice 5.3 · Q4
NCERT Exercise 5.3, Q4	Practice 5.3 · Q5
NCERT Exercise 5.3, Q5	Practice 5.3 · Q6
NCERT Exercise 5.3, Q6	Practice 5.3 · Q7
NCERT Exercise 5.3, Q7	Practice 5.3 · Q8
NCERT Exercise 5.3, Q8	Practice 5.3 · Q9
NCERT Exercise 5.3, Q9	Practice 5.3 · Q10
NCERT Exercise 5.3, Q10	Practice 5.3 · Q11
NCERT Exercise 5.3, Q11	Practice 5.3 · Q12
NCERT Exercise 5.3, Q12	Practice 5.3 · Q13
NCERT Exercise 5.3, Q13	Practice 5.3 · Q14
NCERT Exercise 5.3, Q14	Practice 5.3 · Q15
NCERT Exercise 5.3, Q15	Practice 5.3 · Q16
NCERT Exercise 5.3, Q16	Practice 5.3 · Q17
NCERT Exercise 5.3, Q17	Practice 5.3 · Q18
NCERT Exercise 5.3, Q18	Practice 5.3 · Q19
NCERT Exercise 5.3, Q19	Practice 5.3 · Q20
NCERT Exercise 5.3, Q20	Practice 5.3 · Q21
NCERT Exercise 5.4 (optional), Q1	Practice 5.4 (optional) · Q1
NCERT Exercise 5.4 (optional), Q2	Practice 5.4 (optional) · Q2
NCERT Exercise 5.4 (optional), Q3	Practice 5.4 (optional) · Q3
NCERT Exercise 5.4 (optional), Q4	Practice 5.4 (optional) · Q4
NCERT Exercise 5.4 (optional), Q5	Practice 5.4 (optional) · Q5

Quick check: MCQs

5.1 Introduction2 questions
5.2 Arithmetic Progressions4 questions
5.3 AP ka $n$th Term2 questions
5.3 AP ka $n$th Term3 questions
5.4 AP ke First $n$ Terms ka Sum2 questions
5.4 AP ke First $n$ Terms ka Sum3 questions

Glossary · 7

Key terms (English + Hindi)

arithmetic meanसमांतर माध्यsamaantar maadhya: Do numbers, $a$ aur $c,$ ke theek beech wala number, yaani $\frac{a + c}{2} .$ Chapter mein pehli baar yahan aaya →
arithmetic progressionसमांतर श्रेढ़ीsamaantar shredhi: Aisi list jismein ek term se agle term tak har step par same number judta hai.

common differenceसार्व अंतरsaarv antar: AP ke har step par juda fixed number $d$ : koi bhi term minus uske pehle wala term.

finite APपरिमित समांतर श्रेढ़ीparimit samaantar shredhi: Aisi AP jiska last term ho, isliye usmein terms ki ginti fixed hoti hai.Chapter mein pehli baar yahan aaya →

general termव्यापक पदvyaapak pad: $n$ th term $a_{n} = a + (n - 1) d,$ ek formula jo AP ka har term deta hai.Chapter mein pehli baar yahan aaya →

infinite APअपरिमित समांतर श्रेढ़ीaparimit samaantar shredhi: Aisi AP jo hamesha chalti rehti hai, jiska koi last term nahi.Chapter mein pehli baar yahan aaya →

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

Arithmetic Progressions: Class 10 Maths Chapter 5 Hinglish mein

Is chapter mein aap kya seekhenge

5.1 Introduction

5.2 Arithmetic Progressions

Practice 5.1: answers ke saath

5.3 AP ka $n$ th Term

Practice 5.2: answers ke saath

5.4 AP ke First $n$ Terms ka Sum

Practice 5.3: answers ke saath

Practice 5.4 (optional): answers ke saath

5.5 Summary

Summary aur key points

Board-exam practice

Practice questions answers ke saath (NCERT “Board-exam practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye

5.1 Introduction

5.2 Arithmetic Progressions

5.3 AP ka nth Term

5.4 AP ke First n Terms ka Sum

5.5 Summary

Board-exam practice

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye

5.3 AP ka $n$ th Term

5.4 AP ke First $n$ Terms ka Sum