Question

Find three consecutive terms in A.P. whose sum is 21 and their product is 231.

Answer 1

1. Represent the terms

When dealing with three consecutive terms in A.P., it is simplest to represent them as:

(a – d)

a

(a + d)

where a is the middle term and d is the common difference.

2. Solve for the middle term

Given that the sum of these terms is 21:

(a – d) + a + (a + d) = 21

3a = 21

a = 7

3. Solve for the common difference

Given that the product of these terms is 231 and substituting a = 7:

(7 – d) × 7 × (7 + d) = 231

Divide both sides by 7:

(7 – d)(7 + d) = 33

Using the algebraic identity (x – y)(x + y) = x² – y²:

49 – d² = 33

d² = 49 – 33

d² = 16

d = ±4

4. Determine the terms

If d = 4: The terms are 7 – 4, 7 + 4, which results in 3, 7, 11.

If d = –4: The terms are 7 – (–4), 7 + (–4), which results in 11, 7, 3.

The three consecutive terms in the arithmetic progression are 3, 7 and 11.