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

Chapter Test Questions on Arithmetic and Geometric Progression

ML Aggarwal Class 10 ICSE Maths Solutions Ch-9

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

Solutions of Ch-Test Questions on AP GP

ML Aggarwal Class 10 ICSE Maths Solutions Ch-9

Que-1:  Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.

Ans: First term a = – 5
Common difference d = -3
Then the first four terms are = – 5 + (-3) = -5 – 3 = – 8
-8 + (-3) = – 8 – 3 = -11
– 11 + (-3) = – 11 – 3 = – 14
Therefore, first four terms are -5, -8, -11 and -14.

Que-2: Verify that each of the following lists of numbers is an A.P., and the write its next three terms:

(i) 0, ¼, ½, ¾, …
Ans: First term a = 0
Common difference = ¼ – 0 = ¼
So, next three numbers are ¾ + ¼ = 4/4 = 1
1 + ¼ = (4 + 1)/4 = 5/4
5/4 + ¼ = 6/4 = 3/2
Therefore, the next three term are 1, 5/4 and 3/2.
(ii) 5, 14/3, 13/3, 4, …
Ans: First term a = 5
Common difference = 14/3 – 5 = (14 – 15)/3 = -1/3
So, next three numbers are 4 + (-1/3) = (12 – 1)/3 = 11/3
11/3 + (-1/3) = (11 – 1)/3 = 10/3
10/3 + (-1/3) = (10 – 1)/3 = 9/3 = 3
Therefore, the next three term are 11/3, 10/3 and 3.

Que-3: The n^th term of an A.P. is 6n + 2. Find the common difference.

Ans: n^th term is 6n + 2
So, T_n = 6n + 2
Now, we start giving values, 1, 2, 3, … in the place of n, we get,
T₁ = (6 × 1) + 2 = 6 + 2 = 8
T₂ = (6 × 2) + 2 = 12 + 2 = 14
T₃ = (6 × 3) + 2 = 18 + 2 = 20
T₄ = (6 × 4) + 2 = 24 + 2 = 26
Therefore, A.P. is 8, 14, 20, 26, …
So, common difference d = 14 – 8 = 6

Que-4: Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16^th term and the n^th.

Ans: The first term a = 9
Then, difference d = 12 – 9 = 3
15 – 12 = 3
18 – 15 = 3
Therefore, common difference d = 3
From the formula, a_n = a + (n – 1)d
T_n = a + (n – 1)d
= 9 + (n – 1)3
= 9 + 3n – 3
= 6 + 3n
So, T₁₆ = a + (n – 1)d
= 9 + (16 – 1)3
= 9 + (15)(3)
= 9 + 45
= 54

Que-5: Find the 6^th term from the end of the A.P. 17, 14, 11, …, – 40.

Ans: First term a = 17
Common difference = 14 – 17 = – 3
Last term l = – 40
L = a + (n – 1)d
-40 = 17 + (n – 1)(-3)
-40 – 17 = -3n + 3
– 57 – 3 = -3n
n = -60/-3
n = 20
Therefore, 6^th term form the end = l – (n – 1)d
= – 40 – (6 – 1)(-3)
= – 40 – (5)(-3)
= – 40 + 15
= – 25

Que-6: If the 8^th term of an A.P. is 31 and the 15^th term is 16 more than its 11^th term, then find the A.P.

Ans: a₈ = 31
a₁₅ = the 15^th term is 16 more than its 11^th term = a₁₁ + 16
we know that, a_n = a + (n – 1)d
So, a₈ = a + 7d = 31 … [equation (i)]
a₁₅ = a + 14d = a + 10d + 16
14d – 10d = 16
4d = 16
d = 16/4
d = 4
Now substitute the value of d in equation (i) we get,
a + (7 × 4) = 31
a + 28 = 31
a = 31 – 28
a = 3
So, 3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15
Therefore, A.P. is 3, 7, 11, 15, …

Que-7: The 17^th term of an A.P. is 5 more than twice its 8^th term. If the 11^th term of the A.P. is 43, then find the w^th term.

Ans: a₁₇ = 5 more than twice its 8^th term = 2a₈ + 5
a₁₁ = 43
a_n = ?
We know that, a₁₁ = a + 10d = 43 … [equation (i)]
a₁₇ = 2a₈ + 5
a + 16d = 2(a + 7d) + 5
a + 16d = 2a + 14d + 5
2a – a = 16d – 14d – 5
a = 2d – 5 … [equation (ii)]
Now substitute the value of a in equation (i) we get,
2d – 5 + 10d = 43
12d = 43 + 5
12d = 48
d = 48/12
d = 4
To find out the value of a substitute the value of d in equation (i)
a + (10 × 4) = 43
a + 40 = 43
a = 43 – 40
a = 3
Then, a_n = a + (n – 1)d
= 3 + 4(n – 1)
= 3 + 4n – 4
= 4n – 1

Que-8: The 19^th term of an A.P. is equal to three times its 6^th term. If its 9^th term is 19, find the A.P.

Ans: a₁₉ = 19^th term of an A.P. is equal to three times its 6^th term = 3a₆
a₉ = 19
As we know, a_n = a + (n – 1)d
a₉ = a + 8d = 19 … [equation (i)]
Then, a₁₉ = 3(a + 5d)
a + 18d = 3a + 15d
3a – a = 18d – 15d
2a = 3d
a = (3/2)d
Now substitute the value of a in equation (i) we get,
(3/2)d + 8d = 19
(3d + 16d)/2 = 19
(19/2)d = 19
d = (19 × 2)/19
d = 2
To find out the value of a substitute the value of d in equation (i)
a + 8d = 19
a + (8 × 2) = 19
a + 16 = 19
a = 19 – 16
a = 3
Therefore, A.P. is 3, 5, 7, 9, …

Que-9: If the 3^rd and the 9^th terms of an A.P. are 4 and – 8, respectively, then which term of this A.P. is zero?

Ans:  a₃ = 4
a₉ = – 8
We know that, a₃ = a + 2d = 4 … [equation (i)]
a₉ = a + 8d = -8 … [equation (ii)]
Now, subtracting equation (i) from equation (ii)
(a + 8d) – (a + 2d) = -8 – 4
a + 8d – a – 2d = -12
6d = -12
d = -12/6
d = -2
To find out the value of a substitute the value of d in equation (i)
a + 2d = 4
a + (2 × (-2)) = 4
a – 4 = 4
a = 4 + 4
a = 8
let us assume n^th term be zero, then
a + (n – 1)d = 0
8 + (n – 1)(-2) = 0
-2n + 2 = -8
-2n = -8 – 2
-2n = -10
n = -10/-2
n = 5
Therefore, 0 will be the fifth term.

Que-10: Which term of the list of numbers 5, 2, – 1, – 4, … is – 55?

Ans: First term a = 5
n^th term = -55
Common difference d = 2 – 5 = – 3
We know that, a_n = a + (n – 1)d
– 55 = 5 + (n – 1)(-3)
-55 – 5 = – 3n + 3
-60 – 3 = -3n
-63 = -3n
n = -63/-3
n = 21
Therefore, -55 is the 21^st term.

Que-11: The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is four times its 15^th term.

Ans: The 24^th term of an A.P. is twice its 10^th term = a₂₄ = 2a₁₀
We have to show that, 72^nd term is four times its 15^th term = a₇₂ = ₁₅
We know that, a₂₄ = a + 23d = 2a₁₀
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
2a – a = 23d – 18d
a = 5d … [equation (i)]
a₇₂ = 4a₁₅
a + 71d = 4(a + 14d)
Substitute the value of a we get,
5d + 71d = 4(5d + 14d)
76d = 4(19d)
Therefore, it is proved that 72^nd term is four times its 15^th term.

Que-12: Which term of the list of numbers 20, 19¼, 18½, 17¾, … is the first negative term?

Ans: First term a = 20
Common difference d = 19¼ – 20 = 77/4 – 20 = (77 – 80)/4 = -¾
We know that, a_n = a + (n – 1)d
a_n = 20 + (n – 1) (-¾)
a_n = 20 – ¾n + ¾
a_n = 20 + ¾ – ¾n
a_n = (80 + 3)/4 – ¾n
a_n = 83/4 – ¾n < 0
83/4 < ¾n
83 < 3n
83/3 < n
28 < n
Therefore, 28^th is the first negative term.

Que-13: How many three digit numbers are divisible by 9?

Ans: The three digits numbers which are divisible by 9 are 108, 117, 126, …, 999
Then, first term a = 108
Common difference = 9
Last term = 999
We know that, l = a_n = a + (n – 1)d
999 = 108 + (n – 1)9
999 – 108 = 9n – 9
891 + 9 = 9n
900 = 9n
n = 900/9
n = 100
Therefore, there are 100 three digits numbers.

Que-14: The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.

Ans: The sum of three numbers in A.P. = – 3
The product of three numbers in A.P. = 8
Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d
Now adding 3 numbers = a – d + a + a + d = – 3
3a = -3
a = -3/3
a = -1
From the question, product of 3 numbers is – 35
So, (a – d) × (a) × (a + d) = 8
a(a² – d²) = 8
-1 ((-1)² – d² = 8
1 – d² = 8/-1
1 – d² = -8
d² = 8 + 1
d²= 9
d = √9
d = ±3
Therefore, the numbers are if d = 3 (a – d) = – 1 – 3 = – 4
a = -1
(a + d) = – 1 + 3 = 2
If d = – 6
The numbers are (a – d) = – 1 – (-3) = – 1 + 3 = 2
a = -1
(a + d) = -1 + (-3) = -1 – 3 = -4
Therefore, the numbers -4, -1, 2,… and 2, -1, -4,… are in A.P.

Que-15: The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.

Ans: The angles of a quadrilateral are in A.P.
Greatest angle is double of the smallest angle
Let us assume the greatest angle of the quadrilateral is a + 3d,
Then, the other angles are a + d, a – d, a – 3d
So, a – 3d is the smallest
Therefore, a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
6d + 3d = 2a – a
9d = a … [equation (i)]
We know that the sum of all angles of quadrilateral is 360^o.
a – 3d + a – d + a + d + a+ 3d = 360^o
4a = 360^o
a = 360/4
a = 90^o
Now, substitute the value of a in equation (i) we get,
9d = 90
d = 90/9
d = 10
Substitute the value of a and d in assumed angles,
Greatest angle = a + 3d = 90 + (3 × 10) = 90 + 30 = 120^o
Then, other angles are = a + d = 90^o + 10^o = 100^o
a – d = 90^o – 10^o = 80^o
a – 3d = 90^o – (3 × 10) = 90 – 30 = 60^o
Therefore, the angles of quadrilateral are 120^o, 100^o, 80^o and 60^o.

Que-16: Find the sum of first 20 terms of an A.P. whose n^th term is 15 – 4n.

Ans: n^th term is 15 – 4n
So, a_n = 15 – 4n
Now, we start giving values, 1, 2, 3, … in the place of n, we get,
a₁ = 15 – (4 × 1) = 15 – 4 = 11
a₂ = 15 – (4 × 2) = 15 – 8 = 7
a₃ = 15 – (4 × 3) = 15 – 12 = 3
a₄ = 15 – (4 × 4) = 15 – 16 = – 1
Then, a₂₀ = 15 – (4 × 20) = 15 – 80 = -65
So, 11, 7, 3, -1, … -65 are in A.P.
Therefore, first term a = 11
Common difference = -4
n = 20
S₂₀ = (n/2) [2a + (n – 1)d]
= (20/2) [(2 × 11) + (20 – 1)(-4)]
= 10 [22 – (19) (-4)]
= 10 [22 – 76]
= 10(-54)
= -540
Therefore, the sum of first 20 terms of an A.P. is -540.

Que-17: Find the sum : 18 + 15½ + 13 + … + (-49½)

Ans: First term a = 18
Common difference d = 15½ – 18
= 31/2 – 18
= (31 – 36)/2
= -5/2
Last term = -49½ = -99/2
We know that, a_n = a + (n – 1)d
-99/2 = 18 + (n – 1)(-5/2)
(-99/2) – (18/1) = (n – 1)(-5/2)
(-99 – 36)/2 = (-5/2)(n – 1)
(-135/2) = (-5/2) (n – 1)
(-135/2) × (-2/5) = n – 1
-135/-5 = n – 1
27 = n – 1
n = 27 + 1
n = 28
Then, S_n = (n/2) [2a + (n – 1)d]
S₂₈ = (28/2) [(2 × 18) + (28 – 1)(-5/2)]
S₂₈ = 14[36 + (27 × (-5/2))]
S₂₈ = 14[36 – (135/2)]
S₂₈ = 14 [(72 – 135)/2]
S₂₈ = 14 (-63/2)
S₂₈ = – 441

Que-18: (i) How many terms of the A.P. – 6, (-11/2), – 5, … make the sum – 25?

Ans: Terms of the A.P. is -6, (-11/2) – 5, …
The first term a = -6
Common difference d = (-11/2) – (-6)
= (-11/2) + 6
= (-11 + 12)/2
= ½
The terms are make the sum – 25
Then, S_n = (n/2)(2a + (n – 1)d)
-25 = (n/2) [(2 × (-6)) + (n – 1) (½)]
(-25 × 2) = n [-12 + ½n – ½]
-50 = n [(-25/2) + (½n)]
½n² – (25/2)n + 50 = 0
n² – 25n + 100 = 0
n² – 5n – 20n + 100 = 0
n(n – 5) – 20(n – 5) = 0
(n – 5) (n – 20) = 0
So, n – 5 = 0
n = 5
or n – 20 = 0
n = 20
Therefore, number of terms are 5 or 20.
(ii) Solve the equation 2 + 5 + 8 + … + x = 155
Ans: First term a = 2
Last term = x
Common difference d = 5 – 2 = 3
Then, sum of the terms = 155
L = a + (n – 1)d
x = 2 + (n – 1)3
x = 2 + 3n – 3
x = 3n – 1 … [equation (i)]
We know that, S_n = (n/2) [2a + (n – 1)d]
155 = (n/2) [(2 × 2) + (n – 1) × 3]
155 × 2 = n[4 + 3n – 3]
310 = n(3n + 1)
310 = 3n² + n
3n² + n – 310 = 0
3n² – 30n + 31n – 310 = 0
3n(n – 10) + 31(n – 10) = 0
(n – 10) (3n + 31) = 0
So, n – 10 = 0
n = 10
or 3n + 31 = 0
n = -31/3
negative is not possible.
Therefore, n = 10
Now, substitute the value of n in equation (i),
x = 3n – 1
= (3 × 10) – 1
= 30 – 1
= 29

Que-19: If the third term of an A.P. is 5 and the ratio of its 6^th term to the 10^th term is 7 : 13, then find the sum of first 20 terms of this A.P.

Ans: The third term of an A.P. a₃ = 5
The ratio of its 6^th term to the 10^th term a₆ : a₁₀ = 7 : 13
We know that, a_n = a + (n – 1)d
a₃ = a + (3 – 1)d = 5
= a + 2d = 5 … [equation (i)]
Then, a₆/a₁₀ = 7/13
(a + 5d)/(a + 9d) = 7/13
By cross multiplication we get,
13(a + 5d) = 7(a + 9d)
13a + 75d = 7a + 63d
13a – 7a + 65d – 63d = 0
6a + 2d = 0
Divide by 2 on both side we get,
3a + d = 0
d = -3a … [equation (ii)]
Substitute the value of d in equation (i),
a + 2(-3a) = 5
a – 6a = 5
-5a = 5
a = -5/5
a = -1
Now substitute the value of a in equation (ii),
d = -3(-1)
d = 3
Then, sum of first 20 terms,
= (n/2) [2a + (n – 1)d]
= (20/2)[(2 × (-3)) + (2- – 1)3]
= 10[-2 + 57]
= 10 × 55
= 550

Que-20: The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^th term.

Ans: First term a = 10
The sum of first 14 terms of an A.P. = 1505
25^th term = ?
We know that, S_n = (n/2) [2a + (n – 1)d]
S₁₄ = (n/2) [2a + (n – 1)d]
1505 = (14/2) [(2 × 10) + (14 – 1)d]
1505 = 7[20 + 13d]
1505/7 = 20 + 13d
215 = 20 + 13d
13d = 215 – 20
13d = 195
d = 195/13
d = 15
Then, a_n = a + (n – 1)d
a₂₅ = 10 + (25 – 1)(15)
= 10 + (24)15
= 10 + 360
= 370

Que-21: Find the geometric progression whose 4^th term is 54 and 7^th term is 1458.

Ans: The geometric progression whose 4^th term a₄ = 54
The geometric progression whose 7^th term a₇ = 1458
We know that, a_n = ar^{n – 1}
a₄ = ar^{4 – 1}
a₄ = ar³ = 54
a₇ = ar⁶ = 1458
By dividing both we get,
ar⁶/ar³ = 1458/54
r^{6 – 3} = 27
r³ = 3³
r = 3
To find out a, consider ar³ = 54
a(3)³ = 54
a = 54/27
a = 2
Therefore, a = 2, r = 3
So, G.P. is 2, 6, 18, 54,…

Que-22: The fourth term of a G.P. is the square of its second term and the first term is – 3. Find its 7^th term.

Ans: The fourth term of a G.P. is the square of its second term = a₄ = (a₂)²
The first term a₁ = – 3
We know that, a_n = ar^{n – 1}
a₄ = ar^{4 – 1}
a₄ = ar³
a₂ = ar
Now, ar³ = (ar)²
ar³ = a²r²
r³/r² = a²/a
r^{3 – 2} = a^{2 – 1} … [from a^m/aⁿ = a^{m – n}]
r = a
a₁ = -3
a₇ = ar^{7 – 1}
a₇ = ar⁶
= -3 × (-3)⁶
= -3 × 729
= -2187
Therefore, the 7^th term a₇ = -2187

Que-23: If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.

Ans:  a₄ = x
a₁₀ = y
a₁₆ = z
Now, we have to show that x, y and z are in G.P.
a_n = ar^{n – 1}
a₄ = ar^{4 – 1}
a₄ = ar³ = x
a₁₀ = ar⁹ = y
a₁₆ = ar¹⁵ = z
x, y, z are in G.P.
If y² = xy
Substitute the value of x and y,
y² = (ar⁹)²
y² = a²r¹⁸
Then, xz = ar³ × ar¹⁵
= a^{1 + 1} r^{3 + 15} … [from a^m × aⁿ = a^{m + n}]
= a²r¹⁸
So, y² = xy
Therefore, it is proved that x, y, z are in G.P.

Que-24: How many terms of the G.P. 3, 3/2, ¾ are needed to give the sum 3069/512?

Ans: Sum of the terms S_n = 3069/512
First term a = 3
Common ratio r = (3/2)/3
= (3/2) × (1/3)
= ½
We know that, S_n = a(1 – rⁿ)/(1 – r)
(3069/512) = 3[1 – (½)ⁿ]/ (1 – ½)
(3069/512) = (2 × 3) [1 – (½)ⁿ]
1 – (½)ⁿ = 3069/(512 × 6)
1 – (½)ⁿ = 1023/1024
(½)ⁿ = 1 – (1023/1024)
(½)ⁿ = (1024 – 1023)/1024
(½)ⁿ = 1/1024
(½)ⁿ = (½)¹⁰
By comparing both LHS and RHS,
n = 10
Therefore, there are 10 terms are in the G.P.

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

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