Question

Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP. Hence, find the numbers.

Answer 1

Given: The three numbers 2p – 1, 3p + 1, 11 are in AP.

For three terms to be in AP, the difference between consecutive terms is equal, so 3p + 1 – (2p – 1) = 11 – (3p + 1).

Compute: (3p + 1 – 2p + 1) = p + 2 and (11 – 3p – 1) = 10 – 3p.

Set equal: p + 2 = 10 – 3p

⇒ 4p = 8

⇒ p = 2

Then the numbers are 2p – 1 = 3, 3p + 1 = 7 and 11.

p = 2; the numbers are 3, 7, 11.