AP GP MCQs ML Aggarwal Class 10 ICSE Maths Solutions. Ch-9. We Provide Step by Step Answer of MCQs Questions for Ch-9 Arithmetic and Geometric Progression as council prescribe guideline for upcoming board exam. Visit official Website  CISCE for detail information about ICSE Board Class-10.

Imp MCQs on AP GP With Answer

Board	ICSE
Subject	Maths
Class	10th
Chapter-9	Arithmetic and Geometric Progression
Writer / Book	Understanding
Topics	Solutions of MCQs on AP GP
Academic Session	2025-2026

Solutions of MCQs on AP GP

ML Aggarwal Class 10 ICSE Maths Solutions. Ch-9

Que-1:  In an Arithmetic Progression, if a = 28, d = -4, n = 7, then an is:

(a) 4
(b) 5
(c) 3
(d) 7
Ans: (a) 4
Explanation: For an AP,
an = a+(n-1)d
= 28+(7-1)(-4)
= 28+6(-4)
= 28-24
an=4

Que-2:  If a = 10 and d = 10, then first four terms will be:

(a) 10, 30, 50, 60
(b) 10, 20, 30, 40
(c) 10, 15, 20, 25
(d) 10, 18, 20, 30
Ans: (b) 10, 20, 30, 40
Explanation: a = 10, d = 10
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40

Que-3:  The first term and common difference for the A.P. 3, 1, -1, -3 is:

(a) 1 and 3
(b) -1 and 3
(c) 3 and -2
(d) 2 and 3
Ans: (c) 3 and -2
Explanation: First term, a = 3
Common difference, d = Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2

Que-4:  30th term of the A.P: 10, 7, 4, …, is

(a) 97
(b) 77
(c) -77
(d) -87
Ans: (c) -77
Explanation: Given,
A.P. = 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a1 = 7−10 = −3
As we know, for an A.P.,
an = a +(n−1)d
Putting the values;
a30 = 10+(30−1)(−3)
a30 = 10+(29)(−3)
a30 = 10−87 = −77

Que-5: 11th term of the A.P. -3, -1/2, 2 …. Is

(a) 28
(b) 22
(c) -38
(d) -48
Ans: (b) 22
Explanation: A.P. = -3, -1/2, 2 …
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) -(-3)
⇒(-1/2) + 3 = 5/2
Nth term;
an = a+(n−1)d
a11 = 3+(11-1)(5/2)
a11 = 3+(10)(5/2)
a11 = -3+25
a11 = 22

Que-6: The missing terms in AP: __, 13, __, 3 are:

(a) 11 and 9
(b) 17 and 9
(c) 18 and 8
(d) 18 and 9
Ans: (c)
Explanation: a2 = 13 and
a4 = 3
The nth term of an AP;
an = a+(n−1) d
a2 = a +(2-1)d
13 = a+d ………………. (i)
a4 = a+(4-1)d
3 = a+3d ………….. (ii)
Subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
Now put value of d in equation 1
13 = a+(-5)
a = 18 (first term)
a3 = 18+(3-1)(-5)
= 18+2(-5) = 18-10 = 8 (third term).

Que-7: Which term of the A.P. 3, 8, 13, 18, … is 78?

(a) 12th
(b) 13th
(c) 15th
(d) 16th
Ans: (d) (d) 16th
Explanation: Given, 3, 8, 13, 18, … is the AP.
First term, a = 3
Common difference, d = a2 − a1 = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16

Que-8: The 21st term of AP whose first two terms are -3 and 4 is:

(a) 17
(b) 137
(c) 143
(d) -143
Ans: (b) 137
Explanation: First term = -3 and second term = 4
a = -3
d = 4-a = 4-(-3) = 7
a21=a+(21-1)d
=-3+(20)7
=-3+140
=137

Que-9: If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:

(a) 1
(b) 2
(c) 3
(d) 4
Ans: (a) 1
Explanation: Nth term in AP is:
an = a+(n-1)d
a17 = a+(17−1)d
a17 = a +16d
In the same way,
a10 = a+9d
Given,
a17 − a10 = 7
Therefore,
(a +16d)−(a+9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

Que-10: The number of multiples of 4 between 10 and 250 is:

(a) 50
(b) 40
(c) 60
(d) 30
Ans: (c) 60
Explanation: The multiples of 4 after 10 are:
12, 16, 20, 24, …
So here, a = 12 and d = 4
Now, 250/4 gives remainder 2. Hence, 250 – 2 = 248 is divisible by 2.
12, 16, 20, 24, …, 248
So, nth term, an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60

Que-11: 20th term from the last term of the A.P. 3, 8, 13, …, 253 is:

(a) 147
(b) 151
(c) 154
(d) 158
Ans: (d) 158
Explanation: Given, A.P. is 3, 8, 13, …, 253
Common difference, d= 5.
In reverse order,
253, 248, 243, …, 13, 8, 5
So,
a = 253
d = 248 − 253 = −5
n = 20
By nth term formula,
a20 = a+(20−1)d
a20 = 253+(19)(−5)
a20 = 253−95
a20 = 158

Que-12: The sum of the first five multiples of 3 is:

(a) 45
(b) 55
(c) 65
(d) 75
Ans: (a) 45
Explanation: The first five multiples of 3 is 3, 6, 9, 12 and 15
a=3 and d=3
n=5
Sum, Sn = n/2[2a+(n-1)d]
S5 = 5/2[2(3)+(5-1)3]
=5/2[6+12]
=5/2[18]
=5 x 9
= 45

Que-13: The 10th term of the AP: 5, 8, 11, 14, … is

(a) 32
(b) 35
(c) 38
(d) 185
Ans: (a) 32
Explanation:
Given AP: 5, 8, 11, 14,….
First term = a = 5
Common difference = d = 8 – 5 = 3
nth term of an AP = an = a + (n – 1)d
Now, 10th term = a10 = a + (10 – 1)d
= 5 + 9(3)
= 5 + 27
= 32

Que-14: In an AP, if d = -4, n = 7, an = 4, then a is

(a) 6
(b) 7
(c) 20
(d) 28
Ans: (d) 28
Solution;
Given,
d = -4, n = 7, an = 4
We know that,
an = a + (n – 1)d
4 = a + (7 – 1)(-4)
4 = a + 6(-4)
4 = a – 24
⇒ a = 4 + 24 = 28

Que-15: The list of numbers –10, –6, –2, 2,… is

(a) an AP with d = –16
(b) an AP with d = 4
(c) an AP with d = –4
(d) not an AP
Ans: (b) an AP with d = 4
Explanation:
–10, –6, –2, 2,…
Let a1 = -10, a2 = -6, a3 = -3, a4 = 2
a2 – a1 = -6 – (-10) = 4
a3 – a2 = -2 – (-6) = 4
a4 – a3 = 2 – (-2) = 4
The given list of numbers is an AP with d = 4.

Que-16: If the 2nd term of an AP is 13 and the 5th term is 25, then its 7th term is

(a) 30
(b) 33
(c) 37
(d) 38
Ans: (b) 33
Explanation:
Given,
a2 = 13
a + d = 13
a = 13 – d….(i)
a5 = 25
a + 4d = 25….(ii)
Substituting (i) in (ii),
13 – d + 4d = 25
3d = 12
d = 4
So, a = 13 – 4 = 9
a7 = a + 6d = 9 + 6(4) = 9 + 24 = 33

Que-17: Which term of the AP: 21, 42, 63, 84,… is 210?

(a) 9th
(b) 10th
(c) 11th
(d) 12th
Ans: (b) 10th
Explanation:
Given AP:
21, 42, 63, 84,…
a = 21
d = 42 – 21 = 21
an = 210
a + (n – 1)d = 210
21 + (n – 1)(21) = 210
21 + 21n – 21 = 210
21n = 210
n = 10

Que-18: What is the common difference of an AP in which a18 – a14 = 32?

(a) 8
(b) -8
(c) -4
(d) 4
Ans: (a) 8
Explanation:
Given,
a18 – a14 = 32
We know that, an = a + (n – 1)d
So,
a + 17d – (a + 13d) = 32
17d – 13d = 32
4d = 32
d = 8

Que-19: The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
Ans: (c) Gauss
Explanation:
The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.

Que-20: The sum of first 16 terms of the AP: 10, 6, 2,… is

(a) –320
(b) 320
(c) –352
(d) –400
Ans: (a) -320
Explanation:
Given AP: 10, 6, 2,…
Here, a = 10, d = -4
Sum of first n terms = Sn = (n/2)[2a + (n – 1)d]
The sum of first 16 terms = S16 = (16/2)[2(10) + (16 – 1)(-4)]
= 8[20 + 15(-4)]
= 8(20 – 60)
= 8(-40)
= -320

— : End of AP GP MCQs ML Aggarwal Class 10 ICSE Maths Solutions. of Ch-9 Questions : –

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