Question

Find three numbers in an A.P. whose sum is 21 and the product of the last two numbers is 63.

Answer 1

Let the three numbers in A.P. be a − d, a, a + d

The sum of the three numbers is 21

(a − d) + a + (a + d) = 21

3a = 21

⇒ a = 7

So the three numbers are 7 − d, 7, 7 + d

Product of the last two numbers is 63

Last two numbers are 7 and 7 + d

7(7 + d) = 63

49 + 7d = 63

7d = 63 − 49

7d = 14

⇒ d = 2

= 7 − 2, 7, 7 + 2

= 5, 7, 9

The three numbers in A.P. are 5, 7, and 9.