Question 1

The common difference of the AP $3,7,11,15,\dots$ is:

Accepted Answer

**Answer: (a) $4$.** $d=7-3=4$ (the difference between consecutive terms is constant).

Question 2

The $10$th term of the AP $2,5,8,\dots$ is:

Accepted Answer

**Answer: (a) $29$.** $a=2,\ d=3$; $a_{10}=a+9d=2+9(3)=29$.

Question 3

Which term of the AP $3,8,13,18,\dots$ is $78$?

Accepted Answer

**Answer: (a) $16$th.** $a=3,\ d=5$; $a_n=3+(n-1)5=78\Rightarrow(n-1)5=75\Rightarrow n-1=15\Rightarrow n=16$.

Question 4

If the sum of the first $n$ terms of an AP is $S_n=3n^2+2n$, then its $n$th term is:

Accepted Answer

**Answer: (a) $6n-1$.** $a_n=S_n-S_{n-1}=(3n^2+2n)-\big[3(n-1)^2+2(n-1)\big]=3(2n-1)+2=6n-1$.

Question 5

**Assertion (A):** The $10$th term of the AP $2,4,6,8,\dots$ is $20$. **Reason (R):** The $n$th term of an AP is $a_n=a+(n-1)d$.

Accepted Answer

**Answer: (a)** Using R with $a=2,\ d=2$: $a_{10}=2+9(2)=20$, which is the assertion, so R correctly explains A.

Question 6

Find the $15$th term of the AP $21,18,15,\dots$

Accepted Answer

Here $a=21$ and $d=18-21=-3$. $a_{15}=a+14d=21+14(-3)=21-42=\mathbf{-21}.$

Question 7

Find the sum of the first $20$ terms of the AP $1,3,5,7,\dots$

Accepted Answer

Here $a=1,\ d=2,\ n=20$. $S_{20}=\dfrac{20}{2}\big[2(1)+(20-1)(2)\big]=10\big[2+38\big]=10	imes40=\mathbf{400}.$

Question 8

How many terms of the AP $9,17,25,\dots$ must be taken so that their sum is $636$?

Accepted Answer

Here $a=9,\ d=8$. Using $S_n=636$: $$\frac{n}{2}\big[2(9)+(n-1)8\big]=636\Rightarrow\frac{n}{2}(8n+10)=636.$$ $n(4n+5)=636\Rightarrow 4n^2+5n-636=0.$ $n=\dfrac{-5\pm\sqrt{25+10176}}{8}=\dfrac{-5\pm101}{8}$. Taking the positive value, $n=\dfrac{96}{8}=\mathbf{12}$.

Question 9

The $4$th term of an AP is $11$ and the $8$th term is $23$. Find the AP and its $20$th term.

Accepted Answer

$a+3d=11$ ...(1) and $a+7d=23$ ...(2). Subtracting (1) from (2): $4d=12\Rightarrow d=3$; then $a=11-9=2$. The AP is $\mathbf{2,5,8,11,\dots}$ $a_{20}=a+19d=2+19(3)=\mathbf{59}.$

Question 10

The sum of three numbers in AP is $24$ and their product is $440$. Find the numbers.

Accepted Answer

Let the numbers be $a-d,\ a,\ a+d$. Sum: $(a-d)+a+(a+d)=3a=24\Rightarrow a=8$. Product: $(8-d)(8)(8+d)=440\Rightarrow 8(64-d^2)=440\Rightarrow 64-d^2=55\Rightarrow d^2=9\Rightarrow d=\pm3$. The numbers are $\mathbf{5,\ 8,\ 11}$ (or $11,8,5$).

Question 11

The sum of the first $7$ terms of an AP is $49$ and the sum of the first $17$ terms is $289$. Find the sum of the first $n$ terms.

Accepted Answer

**Using $S_7=49$:** $\dfrac{7}{2}(2a+6d)=49\Rightarrow 2a+6d=14\Rightarrow a+3d=7.$ ...(1) **Using $S_{17}=289$:** $\dfrac{17}{2}(2a+16d)=289\Rightarrow 2a+16d=34\Rightarrow a+8d=17.$ ...(2) Subtract (1) from (2): $5d=10\Rightarrow d=2$; then $a=7-6=1$. $$S_n=\frac{n}{2}\big[2(1)+(n-1)2\big]=\frac{n}{2}(2n)=\mathbf{n^2}.$$

Question 12

A sum of ₹$700$ is to be used to award $7$ cash prizes to students. If each prize is ₹$20$ less than its preceding prize, find the value of each prize.

Accepted Answer

The prizes form an AP with $n=7$, $d=-20$ and sum $S_7=700$. $$S_7=\frac{7}{2}\big[2a+(7-1)(-20)\big]=700.$$ $\dfrac{7}{2}(2a-120)=700\Rightarrow 2a-120=200\Rightarrow 2a=320\Rightarrow a=160.$ Starting at ₹$160$ and decreasing by ₹$20$, the prizes are: $$\mathbf{₹160,\ ₹140,\ ₹120,\ ₹100,\ ₹80,\ ₹60,\ ₹40.}$$ (Check: their sum $=700$.)

Question 13

A cinema hall has seats arranged so that the first row has $20$ seats, the second row $22$ seats, the third row $24$ seats, and so on, the number increasing by a fixed amount in each successive row.

(i) State the common difference of this AP.
(ii) Find the number of seats in the $15$ th row.
(iii) Find the total number of seats in the first $15$ rows.

Accepted Answer

**(i)** The seats are $20,22,24,\dots$, so the common difference is $d=\mathbf{2}$ (with $a=20$). **(ii)** $a_{15}=a+14d=20+14(2)=20+28=\mathbf{48}$ seats. **(iii)** $S_{15}=\dfrac{15}{2}\big[2(20)+(15-1)(2)\big]=\dfrac{15}{2}(40+28)=\dfrac{15}{2}(68)=15	imes34=\mathbf{510}$ seats.

Question 14

Are these Arithmetic Progression important questions free?

Accepted Answer

Yes. All 13 ICSE Class 10 Maths important questions for Arithmetic Progression are free, with full model answers and no login required.

Question 15

Do these Arithmetic Progression questions follow the latest ICSE syllabus?

Accepted Answer

Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Arithmetic Progression important questions?

Accepted Answer

Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.

Question 17

What types of questions are covered for Arithmetic Progression?

Accepted Answer

A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

Arithmetic Progression — Important Questions

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Multiple-choice questions (1 mark)

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Very short answer questions (2 marks)

Short answer questions (3 marks)

Long answer questions (5 marks)

Case-based questions (4 marks)

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