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 17 of 23

1. Q321 CBSE 2020
   Find ​the $11^{\text{th}}$ term from ​the last ​term ​(towards the first term) ​of the ​AP $12,8,4,\dots ,-84$.
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   The $11$th term from the last is $-44$.

   Step-by-step solution →
2. Q322 CBSE 2020
   Solve ​the ​equation: $1+5+9+13+\dots +x=1326$.
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   $x=101$ (with $n=26$ terms)

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3. Q323 CBSE 2020
   Find ​the ​number of numbers ​lying between $146$ and $300$ which are ​divisible ​by both $3$ and $5$.
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   There are $10$ such numbers (150, 165, 180, 195, 210, 225, 240, 255, 270, 285).

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4. Q324 CBSE 2020
   How many ​two-digit numbers ​are ​divisible by ​6 ​?
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   There are $15$ two-digit numbers divisible by $6$.

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5. Q325 CBSE 2020
   In an A.P. ​it ​is ​given that ​common difference is 5 and sum of its first ten terms is ​75. Find the first term of the A.P.
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   First term $a=-15$.

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6. Q326 CBSE 2020
   Find $a$, $b$ and $c$ if ​it ​is given that the ​numbers $a,7,b,23,c$ ​are ​in ​AP.
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   $a=-1,b=15,c=31$

   Step-by-step solution →
7. Q327 CBSE 2020
   If $m$ ​times the $m^{\text{th}}$ ​term of an ​AP is equal to $n$ times its $n^{\text{th}}$ term, ​show that the $(m+n{)}^{\text{th}}$ term of the ​AP is zero.
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   Hence $a_{m+n}=a+(m+n-1)d=0$, i.e. the $(m+n{)}^{\text{th}}$ term is zero.

   Step-by-step solution →
8. Q328 CBSE 2020
   The sum of the first ​30 ​terms ​of ​an A.P. is 1920. ​If the fourth ​term is 18, ​find its $11^{\text{th}}$ ​term.
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   The 11th term is $46$.

   Step-by-step solution →
9. Q329 CBSE 2020
   The ​first and the last terms ​of ​an ​A.P. are $16$ and $136$ respectively. If the common ​difference of the A.P. is $5$, then ​find the ​number of terms in the A.P. ​Also find their sum.
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   Number of terms $n=25$ and sum $S_{25}=1900$.

   Step-by-step solution →
10. Q330 CBSE 2020
    Show that the sum of ​all terms of ​an A.P. ​whose first term ​is $a$, ​the second term ​is $b$ and the last ​term is $c$ ​is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}$.
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    $S=\frac{(a+c)(b+c-2a)}{2(b-a)}$, as required.

    Step-by-step solution →
11. Q331 CBSE 2020
    Solve ​the equation: $1+4+7+10+\dots +x=287$.
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    $x=40$ (the series has 14 terms).

    Step-by-step solution →
12. Q332 CBSE 2020
    For an ​A.P., it is ​given ​that the ​first ​term $(a)=5$, common difference $(d)=3$, and ​the $n^{th}$ term $(a_n)=50$. ​Find $n$ and sum of ​first $n$ terms $(S_n)$ of the A.P.
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    $n=16$ and $S_n=440$.

    Step-by-step solution →
13. Q333 CBSE 2020
    The sum ​of ​four ​consecutive ​numbers in AP ​is $32$ and the ratio of the ​product of the first and last ​terms to the product of two middle terms is $7:15$. ​Find the numbers.
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    The four numbers are $2,6,10,14$ (or $14,10,6,2$ if the AP is taken in decreasing order).

    Step-by-step solution →
14. Q334 CBSE 2020
    The $17^{\text{th}}$ term ​of ​an A.P. is ​5 more than twice its $8^{\text{th}}$ term. ​If $11^{\text{th}}$ term of A.P. ​is ​43; then find its $n^{\text{th}}$ term.
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    $a=3,d=4$, so the $n^{\text{th}}$ term is $a_n=4n-1$.

    Step-by-step solution →
15. Q335 CBSE 2020
    The sum of ​first $6$ terms of an A.P. is $42$. The ratio of its $10^{\text{th}}$ term ​to $30^{\text{th}}$ term is $1:3$. ​Find ​the first and the $13^{\text{th}}$ term ​of the A.P.
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    First term $a=2$ and $13^{\text{th}}$ term $a_{13}=26$.

    Step-by-step solution →
16. Q336 CBSE 2020
    If $4$ ​times ​the $4$th term ​of an AP ​is equal to $18$ ​times ​the $18$th ​term, then ​find the $22$nd term.
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    The $22$nd term is $0$.

    Step-by-step solution →
17. Q337 CBSE 2020
    Find ​the sum of ​all ​odd ​numbers between $100$ ​and $300$.
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    The sum of all odd numbers between $100$ and $300$ is $20000$.

    Step-by-step solution →
18. Q338 CBSE 2020
    How ​many terms of the AP ​: 24, 21, 18, ... must ​be ​taken ​so that their ​sum ​is 78?
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    $n=4$ or $n=13$ terms give a sum of $78$.

    Step-by-step solution →
19. Q339 CBSE 2020
    Solve: $1+4+7+10+\dots +x=287$.
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    $x=40$ (with $n=14$ terms)

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20. Q340 CBSE 2020
    How many terms ​of ​A.P. $3,5,7,9,\dots$ must be ​taken ​to get ​the sum $120$ ?
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    $n=10$ terms must be taken to get the sum 120.

    Step-by-step solution →