ICSE Class 10 Notes on Arithmetic Progression OP Malhotra Maths 2026-27. We Provide Step by Step Answer of all the exercises with Chapter Test of S Chand OP Malhotra Maths . Visit official Website CISCE  for detail information about ICSE Board Class-10.

ICSE Class 10 Notes on Arithmetic Progression OP Malhotra Maths 2026-27

Sequence

A sequence is a set of numbers arranged in a definite order according to some rule. Each number is called a term.

Finite sequence → has a last term. e.g. 4, 9, 14, 19, 24
Infinite sequence → has no last term. e.g. 40, 35, 30, …, −5, …

Series

A series is the sum of the terms of a sequence.

Sequence: 2, 7, 12, 17, …  →  Series: 2 + 7 + 12 + 17 + …

Arithmetic Progression (AP)

A sequence in which terms increase or decrease by the same fixed number is an AP. That fixed number is the common difference (d); the first term is a.
General form of an APa,  a+d,  a+2d,  a+3d, …

AP	a	d
5, 8, 11, 14, 17	5	3
9, 5, 1, −3, −7	9	−4

👉 d > 0 → AP is increasing  |  d < 0 → AP is decreasing.

nth Term Formula

nth term of an AP ⇒ T_n = a + (n − 1)d
Common difference ⇒ d = T_n − T_n−1

Solved Examples

Ex. 1 Find the 16th term of the AP 3, 10, 17, 24, …

Solution: a = 3, d = 10 − 3 = 7, n = 16

T_n = a + (n−1)d
T₁₆ = 3 + (16−1)×7 = 3 + 105

T₁₆ = 108

Ex. 2 The 7th term of an AP is −4 and its 13th term is −16. Find the AP.

Solution: Let a = first term, d = common difference.

T₇ = a + 6d = −4  …(i)
T₁₃ = a + 12d = −16  …(ii)

Subtracting (i) from (ii):

6d = −12  ⟹  d = −2
From (i): a + 6(−2) = −4  ⟹  a = 8

The required AP is 8, 6, 4, 2, …

Ex. 3 Check whether −83 is a term of the AP 5, 2, −1, −4, …

Solution: a = 5, d = 2 − 5 = −3. Let −83 be the nth term:

T_n = a + (n−1)d = −83
5 + (n−1)(−3) = −83
−3(n−1) = −88  ⟹  n−1 = 88/3  ⟹  n = 92/3

n is not a whole number → −83 is NOT a term of the given AP.

Ex. 4 Find the 12th term from the last term of the AP 21, 18, 15, …, −81.

Solution: a = 21, d = −3, last term T_n = −81. First find total terms:

−81 = 21 + (n−1)(−3)
−3(n−1) = −102  ⟹  n − 1 = 34  ⟹  n = 35

12th from the end = (35 − 12 + 1)th from start = 24th term

T₂₄ = 21 + (24−1)(−3) = 21 − 69

T₂₄ = −48

Ex. 5 Find the middle term of the AP 6, 13, 20, …, 216.

Solution: a = 6, d = 7, last term T_n = 216. Find total terms:

216 = 6 + (n−1)×7  ⟹  7(n−1) = 210  ⟹  n = 31

n = 31 (odd) → middle term = 16th term

T₁₆ = 6 + (16−1)×7 = 6 + 105

Middle term = 111

Practice Questions on this topic : Exercise- 9(a)

Sum of n terms

Sum of n terms of AP (S) = n/2[2a + (n – 1)d]

or

S = n/2(a+l)

where n is number of terms ,
a is the first term of AP
l is the last term of  AP

Note: 

• Each of the formulas contains four quantities . Thus three being known , fourth can be found out.
• Also, T_n = S_n+ S_n-1

Practice Questions on this topic : Exercise-9(b)

GEOMETRIC PROGRESSION (G.P.)

A Geometric Progression (G.P.) is a sequence in which each term is obtained
by multiplying the previous term by the same number.

This fixed number is called the Common Ratio (r).

Examples

• 2, 4, 8, 16, …
• 27, 9, 3, 1, …
• 1, -2, 4, -8, …

Common Ratio (r)

r = Next Term/Previous Term

Example
2, 6, 18, 54
r = 6 ÷ 2 = 3

Standard Form of G.P.

a, ar, ar², ar³, ar⁴, …

• a = First term
• r = Common ratio

nth Term of G.P.

• a = First term
• r = Common ratio
• n = Term number

Solved Examples

Example 1 : Check whether 27, 9, 3, 1 is a G.P.

Solution :9 ÷ 27 = 1/3
3 ÷ 9 = 1/3
1 ÷ 3 = 1/3

Answer: Yes, it is a G.P.

Example 2 : Find the common ratio of 2, 6, 18, 54.

Solution :
r = 6 ÷ 2 = 3

Answer: r = 3

Example 3 : Find the 5th term of 3, 6, 12, 24, …

Solution :a = 3, r = 2
T₅ = 3 × 2⁴ = 48

Answer: 48

Example 4 : Which term of the G.P. 6, 2, 2/3, … is 2/243?
Solution:
a = 6, r = 1/3
6 × (1/3)ⁿ⁻¹ = 2/243
(1/3)⁶ = (1/3)ⁿ⁻¹
n = 7

Answer: 7th term

Practice Questions on this topic : Exercise-9(c)

SUM OF n TERMS OF A GEOMETRIC PROGRESSION

The sum of the first n terms of a Geometric Progression is denoted by S_n.

If the first term is a, common ratio is r and number of terms is n, then

S_n = a + ar + ar² + ar³ + … + arⁿ⁻¹

Formulae

Case 1 : When r < 1

S_n = a(1 − rⁿ)/(1 − r)

Case 2 : When r > 1

S_n = a(rⁿ − 1)/(r − 1)

Note: Both formulae give the same answer. Use the one which is easier.

Solved Examples

Example 1 : Find the sum of the first 5 terms of 2, 4, 8, 16, …

Solution:

Given:
a = 2,
r = 2,
n = 5

S₅ = 2(2⁵ − 1)/(2 − 1)
= 62

Answer: 62

Example 2 : Find the sum of the first 4 terms of 81, 27, 9, 3, …

Solution :

Given:
a = 81,
r = 1/3,
n = 4

S₄ = 81(1 − (1/3)⁴)/(1 − 1/3)
= 120

Answer: 120

Example 3 : How many terms of the G.P. 5, 20, 80, … have sum 6825?

Solution :
Given:
a = 5,
r = 4,
S_n = 6825

6825 = 5(4ⁿ − 1)/3
4ⁿ = 4096
4⁶ = 4096

Answer: n = 6

Example 4 : The third term of a G.P. is 5/2 and the eighth term is 5/64.
Find the sum of the first 10 terms.

Solution :
T₃ = ar² = 5/2
T₈ = ar⁷ = 5/64
r = 1/2
a = 10

S₁₀ = 10(1 − (1/2)¹⁰)/(1 − 1/2)
= 5115/256

Answer: 5115/256

Practice Questions on this topic : Exercise-9(d)

Self Evaluation and Revision : Coming Soon

— : End of ICSE Class 10 Notes on Arithmetic Progression OP Malhotra Maths 2026-27 :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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