Class 10 Maths · Chapter 5

Arithmetic Progressions

441 Class 10 Maths questions on Arithmetic Progressions from the NCERT textbook, NCERT Exemplar and CBSE board papers 2015–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 21 of 23

1. Q401 CBSE 2016
   Write ​the $11^{\text{th}}$ ​term ​of ​the following ​A.P. ​: $3,\frac{1}{2},-2,\dots$
   Show answer

   The 11th term is $t_{11}=-22$.

   Step-by-step solution →
2. Q402 CBSE 2016
   For what ​value ​of $k$ ​will $k+9$, $2k-1$ and $2k+7$ ​be the ​consecutive terms of ​an A.P.?
   Show answer

   $k=18$.

   Step-by-step solution →
3. Q403 CBSE 2016
   For what value ​of $k$ ​will $k+9$, $2k-1$ and $2k+7$ are ​the ​consecutive ​terms of ​an A.P.?
   Show answer

   $k=18$.

   Step-by-step solution →
4. Q404 CBSE 2016
   For what value ​of $k$ will the consecutive terms $2k+1$, $3k+3$ ​and $5k-1$ ​form ​an ​A.P. ?
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   $k=6$.

   Step-by-step solution →
5. Q405 CBSE 2016
   Find the 9th ​term from the ​end ​(towards ​the ​first term) of the A.P. $5,9,13,\dots ,185$.
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   The 9th term from the end is $153$.

   Step-by-step solution →
6. Q406 CBSE 2016
   For ​what ​value ​of $k$ will $k+9$, $2k-1$ and $2k+7$ are ​the ​consecutive terms of an ​A.P. ?
   Show answer

   $k=18$.

   Step-by-step solution →
7. Q407 CBSE 2016
   If ​the sum of ​first $9$ ​terms of ​an AP is ​equal to the sum of ​its first $11$ terms, then find the sum of its ​first $20$ terms.
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   Sum of the first $20$ terms, $S_{20}=0$.

   Step-by-step solution →
8. Q408 CBSE 2016
   The ​4th term ​of an ​A.P. is zero. Prove that the 25th term ​of ​the A.P. is three times its ​11th term.
   Show answer

   $a_{25}=21d$ and $a_{11}=7d$, so $a_{25}=3a_{11}$, as required.

   Step-by-step solution →
9. Q409 CBSE 2016
   Find ​how ​many two ​digit ​numbers are ​divisible ​by $4$.
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   There are $22$ two-digit numbers divisible by $4$.

   Step-by-step solution →
10. Q410 CBSE 2016
    The $4^{\text{th}}$ term of an ​A.P. ​is ​zero. Prove that ​the $25^{\text{th}}$ ​term of ​the A.P. is three times its $11^{\text{th}}$ term.
    Show answer

    Since $a_4=0\Rightarrow a=-3d$, we get $a_{25}=21d$ and $a_{11}=7d$, hence $a_{25}=3a_{11}$. Proved.

    Step-by-step solution →
11. Q411 CBSE 2016
    If the ​ratio ​of sum of the first $m$ and $n$ terms of an ​A.P. is $m^2:n^2$, show that the ​ratio ​of its $m^{th}$ ​and $n^{th}$ terms is $(2m-1):(2n-1)$.
    Show answer

    $\frac{a_m}{a_n}=\frac{2m-1}{2n-1}$, as required.

    Step-by-step solution →
12. Q412 CBSE 2016
    The $4^{th}$ term of an A.P. ​is ​zero. Prove ​that the $25^{th}$ term of ​the A.P. ​is three times ​its $11^{th}$ ​term.
    Show answer

    $a_{25}=21d=3\times 7d=3a_{11}$, hence the 25th term is three times the 11th term (proved).

    Step-by-step solution →
13. Q413 CBSE 2016
    How many terms of the A.P. $18,16,14,\dots$ ​be taken ​so that ​their ​sum ​is zero?
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    Taking $19$ terms of the A.P. Makes their sum zero.

    Step-by-step solution →
14. Q414 CBSE 2016
    Divide $207$ ​into ​three parts such that ​these are ​in ​A.P. and ​the product of the two smaller ​parts is $4623$.
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    The three parts are $67$, $69$ and $71$.

    Step-by-step solution →
15. Q415 CBSE 2016
    Find the ​sum of ​first $40$ ​positive integers ​which ​give ​a remainder $1$ ​when divided by $6$.
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    The required sum is $4720$.

    Step-by-step solution →
16. Q416 CBSE 2016
    Divide $56$ in four ​parts in A.P. ​such that the ratio of the ​product of their ​extremes ​($1^{st}$ and $4^{th}$) ​to the product of ​means ​($2^{nd}$ and $3^{rd}$) is $5:6$.
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    The four parts are $8,12,16,20$.

    Step-by-step solution →
17. Q417 CBSE 2016
    If ​the ratio of the sum of first $n$ ​terms ​of ​two ​A.P.'s ​is $(7n+1):(4n+27)$, find the ratio of ​their $m^{th}$ terms.
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    The ratio of the $m^{th}$ terms is $(14m-6):(8m+23)$.

    Step-by-step solution →
18. Q418 CBSE 2016
    If the ​sum of first $7$ terms ​of an A.P. is $49$ ​and that of its first $17$ ​terms is $289$, ​find the ​sum of first $n$ ​terms of ​the A.P.
    Show answer

    $S_n=n^2$.

    Step-by-step solution →
19. Q419 CBSE 2016
    The digits ​of a ​positive number of three digits ​are in A.P. and their sum is $15$. The number obtained by reversing the digits is $594$ less ​than the ​original ​number. Find ​the number.
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    The number is $852$.

    Step-by-step solution →
20. Q420 CBSE 2016
    The sums of first $n$ terms ​of three arithmetic ​progressions are $S_1$, $S_2$ and $S_3$ respectively. The first term ​of each ​A.P. is $1$ and ​their common differences ​are $1$, $2$ and $3$ respectively. Prove that $S_1+S_3=2S_2$.
    Show answer

    $S_1+S_3=\frac{n}{2}[4+4(n-1)]=2S_2$. Proved.

    Step-by-step solution →