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 20 of 23

1. Q381 CBSE 2017
   Find ​how ​many ​integers ​between $200$ ​and $500$ are ​divisible ​by $8$.
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   There are $37$ integers between $200$ and $500$ that are divisible by $8$.

   Step-by-step solution →
2. Q382 CBSE 2017
   Find the value ​of $p$ for ​which the numbers $2p-1$, $3p+1$, $11$ are in ​A.P. ​Hence find ​the numbers.
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   $p=2$, and the numbers are $3,7,11$ (common difference $4$).

   Step-by-step solution →
3. Q383 CBSE 2017
   For what value ​of $n$, ​are ​the $n^{th}$ terms ​of two ​A.Ps $63,65,67,\dots$ ​and $3,10,17,\dots$ ​equal?
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   $n=13$.

   Step-by-step solution →
4. Q384 CBSE 2017
   Which term of ​the ​A.P. $8,14,20,26,\dots$ will ​be $72$ more ​than ​its $41^{\text{st}}$ ​term?
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   The $53^{\text{rd}}$ term ($n=53$)

   Step-by-step solution →
5. Q385 CBSE 2017
   If $m^{th}$ ​term of an A.P. ​is $\frac{1}{n}$ ​and $n^{th}$ ​term ​is $\frac{1}{m}$, then find the sum ​of ​its first $mn$ terms.
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   The sum of the first $mn$ terms is $\frac{mn+1}{2}$.

   Step-by-step solution →
6. Q386 CBSE 2017
   Find the ​sum of $n$ ​terms ​of the ​series $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\cdots$
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   $S_n=\frac{7n-1}{2}$

   Step-by-step solution →
7. Q387 CBSE 2017
   If ​the ​sixth term of ​an A.P. is ​zero, ​then show that its $33^{\text{rd}}$ term ​is three times its $15^{\text{th}}$ term.
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   Since $a_{33}=27d$ and $a_{15}=9d$, we have $a_{33}=3a_{15}$, as required.

   Step-by-step solution →
8. Q388 CBSE 2017
   The first ​term ​of an A.P. is $5$, the last term is $45$ and ​the sum of all its terms is $400$. Find the number of ​terms and the common ​difference of ​the A.P.
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   Number of terms $n=16$ and common difference $d=\frac{8}{3}$.

   Step-by-step solution →
9. Q389 CBSE 2017
   Find ​the ​sum of ​the following ​series: $5+(-41)+9+(-39)+13+(-37)+17+\dots +(-5)+81+(-3)$.
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   The sum of the series is $420$.

   Step-by-step solution →
10. Q390 CBSE 2017
    If ​the $10^{\text{th}}$ ​term ​of an ​A.P. is $52$ and the $17^{\text{th}}$ term is $20$ more than the $13^{\text{th}}$ ​term, ​find the A.P.
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    The A.P. Is $7,12,17,22,\dots$ (first term $a=7$, common difference $d=5$).

    Step-by-step solution →
11. Q391 CBSE 2017
    How many terms ​of ​an ​A.P. $9,17,25,\dots$ ​must be ​taken ​to give a sum of $636$?
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    $12$ terms of the A.P. Must be taken to give a sum of $636$.

    Step-by-step solution →
12. Q392 CBSE 2017
    A sum of ₹ 4,250 is to be used to give 10 cash prizes ​to students ​of a ​school ​for their overall academic performance. If each prize is ₹ 50 ​less than its preceding prize, find ​the value of each of the prizes. What values are shown by the ​school in giving these prizes?
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    The prizes (in ₹) are $650,600,550,500,450,400,350,300,250,200$. The school shows the value of encouraging and rewarding academic excellence / merit.

    Step-by-step solution →
13. Q393 CBSE 2017
    If $1+4+7+10+\dots +x=287$, ​find the ​value ​of $x$.
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    $x=40$ (the series has $14$ terms).

    Step-by-step solution →
14. Q394 CBSE 2017
    If the ratio ​of the sum of ​the first $n$ ​terms ​of ​two A.Ps is $(7n+1):(4n+27)$, then find the ratio of ​their $9^{\text{th}}$ ​terms.
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    The ratio of the 9th terms is $24:19$.

    Step-by-step solution →
15. Q395 CBSE 2017
    The ​ratio of the sums of first $m$ ​and first $n$ 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

    Since the sum condition forces $d=2a$, the $m$th term is $a(2m-1)$ and the $n$th term is $a(2n-1)$, so their ratio is $(2m-1):(2n-1)$, as required.

    Step-by-step solution →
16. Q396 CBSE 2017
    In ​a construction ​work, a contractor has to pay penalty for delay ​beyond a ​certain date as follows : $\text{₹}200$ ​for ​the first day, $\text{₹}250$ for the second day, $\text{₹}300$ for the third day and ​so on. How much money does the contractor have to pay as penalty, ​if he has delayed the work by $10$ days ? Why is completing any work on time a must ?
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    The contractor must pay $\text{₹}4250$ as penalty. Completing work on time avoids growing penalties and financial loss and reflects discipline and responsibility.

    Step-by-step solution →
17. Q397 CBSE 2017
    The sum ​of the $4^{th}$ ​and the $8^{th}$ terms of an ​AP ​is $24$ and the sum of ​the $6^{th}$ ​and the $10^{th}$ terms is $44$. ​Find the ​sum of the first $10$ terms of the AP.
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    The sum of the first $10$ terms is $S_{10}=95$.

    Step-by-step solution →
18. Q398 CBSE 2017
    If the ​ratio of the sum of the first $n$ terms ​of two A.Ps ​is $(7n+1):(4n+27)$, ​then ​find ​the ratio of their $9^{th}$ ​terms.
    Show answer

    The ratio of the 9th terms is $24:19$.

    Step-by-step solution →
19. Q399 CBSE 2017
    If the ratio of the ​sum of ​the first $n$ terms ​of two A.P.s ​is $(7n+1):(4n+27)$, then find the ratio ​of ​their $9^{\text{th}}$ terms.
    Show answer

    The ratio of the 9th terms is $24:19$.

    Step-by-step solution →
20. Q400 CBSE 2017
    A child puts one ​five-rupee coin of ​her saving in the piggy ​bank on the first day. She increases her ​saving by one five-rupee coin daily. If the piggy bank can hold $190$ ​coins of five rupees in all, find the number ​of days she can continue to put the five-rupee coins ​into it and find the total money she saved. Write your views on the habit of saving.
    Show answer

    She can add coins for $19$ days, saving a total of ₹$950$. (View: regular daily saving, however small, accumulates into a large sum and builds financial discipline.)

    Step-by-step solution →