Arithmetic Progression

Q: Three numbers in AP have sum 15 and product 80. Find them.

Take a - d, a, a + d. Then 3a = 15, so a = 5, and 5(25 - d2) = 80 gives d = ± 3. The numbers are 2, 5, 8.

ICSE Class 10 · Mathematics · Chapter 10

What this chapter covers

Sequences and arithmetic progressions
The general term of an arithmetic progression
Using the general term to solve problems
Three terms in an AP and the middle-term property
Sum of the first n terms of an arithmetic progression
Using the sum formula to find unknowns
Simple applications of arithmetic progression

Key points

A sequence is an ordered list of numbers; an arithmetic progression is a sequence in which the same number d is added each time.
The common difference is d = a_n+1 - a_n; it may be positive, negative, zero or fractional, and needs at least three terms to test.
General term: a_n = a + (n - 1)d, a linear expression in n whose coefficient of n is d.
A number is a term of an AP only if solving a_n = that number gives a natural number n.
For two terms of the same AP, a_m - a_k = (m - k)d.
The nth term from the end of a finite AP with last term l is l - (n - 1)d.
Three numbers a, b, c are in AP exactly when 2b = a + c; take unknown terms as a - d, a, a + d.
Sum of the first n terms: S_n = (n)/(2)[2a + (n - 1)d] = (n)/(2)(a + l).
Sum of the first n natural numbers: (n(n + 1))/(2).
a_n = S_n - S_n-1 for n ≥ 2, and a₁ = S₁.
When n is unknown in a sum problem, solve the quadratic and keep only natural-number values.
In applications, decide first whether the question asks for one term (a_n) or a total (S_n), and watch the units.

Questions answered

Define an arithmetic progression and its common difference.

An arithmetic progression is a sequence in which every term after the first is obtained by adding the same fixed number to the term before it. The common difference is that fixed number, equal to any term minus the term just before it.

Find the common difference of the AP 2.5, 1, -0.5, -2, …

d = 1 - 2.5 = -1.5.

Find the 15th term of the AP -4, -1, 2, 5, …

a₁₅ = -4 + 14 × 3 = 38.

Which term of the AP 64, 60, 56, … is 0?

64 + (n - 1)(-4) = 0 gives n - 1 = 16, so 0 is the 17th term.

Is 100 a term of the AP 3, 7, 11, …? Give a reason.

3 + 4(n - 1) = 100 gives n - 1 = (97)/(4), which is not a whole number, so 100 is not a term.

The 3rd term of an AP is 16 and its 7th term exceeds its 5th term by 12. Write the first four terms.

a₇ - a₅ = 2d = 12, so d = 6; a + 2d = 16 gives a = 4. The AP is 4, 10, 16, 22, …

Find the 8th term from the end of the AP 3, 8, 13, …, 253.

253 - 7 × 5 = 218.

Three numbers in AP have sum 15 and product 80. Find them.

Take a - d, a, a + d. Then 3a = 15, so a = 5, and 5(25 - d²) = 80 gives d = ± 3. The numbers are 2, 5, 8.

Past board exam questions

ICSE 2027 sample paper · 4 marks

An Arithmetic Progression (A.P.) consists of all whole numbers which are divisible by 3 and 5. (a) Write its first term and the common difference. (b) Find the 10th term of the A.P. (c) Find the sum of the first 10 terms of the A.P.

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Numbers divisible by both 3 and 5 are the multiples of 15: 15, 30, 45, … (a) First term a = 15, common difference d = 15. (b) T₁₀ = a + 9d = 15 + 9 × 15 = 150. (c) S₁₀ = (10)/(2)[2 × 15 + 9 × 15] = 5 × 165 = 825. (Answer as in the board's marking scheme, which starts the A.P. at 15.)

ICSE 2026 sample paper · 4 marks

-11, -7, -3, ………….,49, 53 are the terms of a progression. Answer the following: (a) What is the type of progression? (b) How many terms are there in all? (c) Calculate the value of middle most term.

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(a) −7 − (−11) = 4 and −3 − (−7) = 4: the difference is constant, so it is an Arithmetic Progression (A.P.) with a = −11, d = 4. (b) a + (n - 1)d = 53 ⇒ -11 + 4(n - 1) = 53 ⇒ n - 1 = 16 ⇒ n = 17. (c) The middle term is the (17 + 1)/(2) = 9th term: T₉ = -11 + 8 × 4 = 21.

ICSE 2026 board paper · 4 marks

The fourth and seventh terms of an Arithmetic Progression (A.P.), are 60 and 114 respectively. Find the: (a) first term and common difference. (b) sum of its first 10 terms.

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(a) a + 3d = 60 and a + 6d = 114. Subtracting, 3d = 54 ⇒ d = 18; a = 60 - 54 = 6. First term = 6, common difference = 18. (b) S₁₀ = (10)/(2)[2(6) + 9(18)] = 5(12 + 162) = 5 × 174 = 870.

ICSE 2026 sample paper · 3 marks

An Arithmetic Progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.

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a = 3. S₈ = 2S₅: (8)/(2)[2 × 3 + 7d] = 2 × (5)/(2)[2 × 3 + 4d] 4(6 + 7d) = 5(6 + 4d) ⇒ 24 + 28d = 30 + 20d ⇒ 8d = 6 ⇒ d = (3)/(4).

ICSE 2025 board paper · 3 marks

The first term of an Arithmetic Progression (A.P.) is 5, the last term is 50 and their sum is 440. Find: (a) the number of terms (b) common difference

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(a) S_n = (n)/(2)(a + l) ⇒ 440 = (n)/(2)(5 + 50) ⇒ n = (880)/(55) = 16. (b) l = a + (n - 1)d ⇒ 50 = 5 + 15d ⇒ d = 3.

ICSE 2025 compartment paper · 3 marks

164, 160, 156, 152, …… are in Arithmetic Progression (A.P.). Find: (a) which term is equal to 0. (b) the sum of its first 20 terms.

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a = 164, d = -4. (a) a_n = 164 + (n - 1)(-4) = 0 ⇒ n - 1 = 41 ⇒ n = 42. The 42nd term is 0. (b) S₂₀ = (20)/(2)[2(164) + 19(-4)] = 10(328 - 76) = 2520.

ICSE 2024 sample paper · 1 mark

Which of the following is/are an Arithmetic Progression (A.P.)? 1. 1, 4, 9, 16,………. 2. √(3), 2√(3), 3√(3), 4√(3),……… 3. 8, 6, 4, 2,………

only 1.
only 2.
only 2. and 3.
all 1., 2. and 3.

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only 2. and 3.

ICSE 2024 board paper · 1 mark

The 7th term of the given Arithmetic Progression (A.P.): (1)/(a), ((1)/(a) + 1), ((1)/(a) + 2) ldots is:

((1)/(a) + 6)
((1)/(a) + 7)
((1)/(a) + 8)
((1)/(a) + 7⁷)

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((1)/(a) + 6)