Class 10 Maths · Chapter 5

Arithmetic Progressions

152 Class 10 Maths questions on Arithmetic Progressions from the NCERT textbook, NCERT Exemplar and CBSE board papers 2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 5 of 8

1. Q81 Exemplar · Exercise 5.3
   The sum ​of the ​5th and the 7th terms of an ​AP ​is 52 ​and the 10th ​term is 46. Find the AP.
   Show answer

   The AP is $1,6,11,16,21,\dots$ with first term $a=1$ and common difference $d=5$.

   Step-by-step solution →
2. Q82 Exemplar · Exercise 5.3
   Find ​the ​20th term ​of the AP whose 7th term is ​24 less than ​the 11th ​term, first term ​being 12.
   Show answer

   The 20th term is $a_{20}=126$.

   Step-by-step solution →
3. Q83 Exemplar · Exercise 5.3
   If the $9^{th}$ term ​of an AP ​is ​zero, prove ​that ​its $29^{th}$ ​term ​is ​twice its $19^{th}$ term.
   Show answer

   Proved: with $a_9=0$ we get $a=-8d$, so $a_{19}=10d$ and $a_{29}=20d=2a_{19}$.

   Step-by-step solution →
4. Q84 Exemplar · Exercise 5.3
   Find whether ​55 ​is a term ​of the AP: 7, 10, ​13, ... or ​not. If yes, find which ​term it ​is.
   Show answer

   Yes, 55 is a term of the AP -- it is the 17th term.

   Step-by-step solution →
5. Q85 Exemplar · Exercise 5.3
   Determine $k$ so ​that $k^2+4k+8$, $2k^2+3k+6$, $3k^2+4k+4$ ​are ​three ​consecutive ​terms of ​an ​AP.
   Show answer

   $k=0$; the three terms become $8$, $6$, $4$ with common difference $-2$.

   Step-by-step solution →
6. Q86 Exemplar · Exercise 5.3
   Split ​207 ​into three parts such ​that these ​are in AP ​and the product ​of the two smaller parts ​is 4623.
   Show answer

   The three parts are 67, 69 and 71 (an AP with common difference 2).

   Step-by-step solution →
7. Q87 Exemplar · Exercise 5.3
   The angles of a triangle are in AP. ​The ​greatest angle ​is twice ​the least. Find ​all the angles of the ​triangle.
   Show answer

   The angles are $40^{\circ },60^{\circ }$ and $80^{\circ }$.

   Step-by-step solution →
8. Q88 Exemplar · Exercise 5.3
   If the $n$th terms of the ​two ​APs: $9,7,5,\dots$ and $24,21,18,\dots$ are ​the ​same, find the ​value of $n$. ​Also ​find ​that term.
   Show answer

   $n=16$, and the common term is $-21$.

   Step-by-step solution →
9. Q89 Exemplar · Exercise 5.3
   If sum ​of the 3rd ​and ​the 8th terms ​of an AP is 7 and the sum ​of the 7th and ​the 14th terms ​is $-3$, find the 10th term.
   Show answer

   The 10th term is $a_{10}=-1$ (with $a=8$, $d=-1$).

   Step-by-step solution →
10. Q90 Exemplar · Exercise 5.3
    Find ​the $12^{\text{th}}$ ​term ​from the ​end ​of ​the AP: $-2,-4,-6,\dots ,-100$.
    Show answer

    The 12th term from the end is $-78$.

    Step-by-step solution →
11. Q91 Exemplar · Exercise 5.3
    Which term ​of the AP: 53, ​48, 43, ​... ​is the first negative ​term?
    Show answer

    The 12th term is the first negative term ($a_{12}=-2$).

    Step-by-step solution →
12. Q92 Exemplar · Exercise 5.3
    How many numbers lie between ​10 and 300, which when ​divided ​by 4 ​leave a remainder ​3?
    Show answer

    There are $73$ such numbers.

    Step-by-step solution →
13. Q93 Exemplar · Exercise 5.3
    Find ​the ​sum of ​the two middle most ​terms of ​the ​AP: $-\frac{4}{3},-1,-\frac{2}{3},\dots ,4\frac{1}{3}$.
    Show answer

    The sum of the two middle-most terms is $3$.

    Step-by-step solution →
14. Q94 Exemplar · Exercise 5.3
    The first ​term of ​an AP is $-5$ and the ​last term is $45$. If the sum of ​the terms ​of the AP is $120$, then find the number ​of terms and ​the common difference.
    Show answer

    The AP has $n=6$ terms with common difference $d=10$.

    Step-by-step solution →
15. Q95 Exemplar · Exercise 5.3
    Find the sum: ​

    1. (i)$1+(-2)+(-5)+(-8)+\dots +(-236)$ ​
    2. (ii)$\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\dots$ upto $n$ ​terms ​
    3. (iii)$\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+\dots$ ​to 11 terms.
    Show answer

    (i) $-9400$; (ii) $\frac{7n-1}{2}$; (iii) $\frac{11(11a-6b)}{a+b}$.

    Step-by-step solution →
16. Q96 Exemplar · Exercise 5.3
    Which ​term of ​the AP: $-2,-7,-12,\dots$ will be $-77$? Find ​the sum of this ​AP upto the term $-77$.
    Show answer

    $-77$ is the $16$th term, and the sum up to it is $S_{16}=-632$.

    Step-by-step solution →
17. Q97 Exemplar · Exercise 5.3
    If $a_n=3-4n$, show that $a_1,a_2,a_3,\dots$ ​form ​an AP. ​Also find $S_{20}$.
    Show answer

    The sequence is an AP with $a=-1$, $d=-4$; $S_{20}=-780$.

    Step-by-step solution →
18. Q98 Exemplar · Exercise 5.3
    In ​an ​AP, ​if $S_n=n(4n+1)$, ​find ​the AP.
    Show answer

    The AP is $5,13,21,29,\dots$ with first term $5$ and common difference $8$ ($a_n=8n-3$).

    Step-by-step solution →
19. Q99 Exemplar · Exercise 5.3
    In ​an ​AP, if $S_n=3n^2+5n$ ​and $a_k=164$, ​find the value of $k$.
    Show answer

    $k=27$.

    Step-by-step solution →
20. Q100 Exemplar · Exercise 5.3
    If $S_n$ denotes the sum ​of ​first $n$ ​terms ​of an ​AP, ​prove ​that $S_{12}=3(S_8-S_4)$.
    Show answer

    Both sides equal $12a+66d$, hence $S_{12}=3(S_8-S_4)$ is proved.

    Step-by-step solution →