Question

the sum of three numbers in AP is 21 . The product of the first and the third number exceeds the second number by 6. Find those numbers.

Answer 1

Let the AP be a-d,a,a+d

Given:

a-d+a+a+d=21

3a=21

=a=21/3

=a=7

Product of first and third number:

(a+d)(a-d)

=a²-d²

a²-d² exceeds second number by 6.

=a²-d²-a=6

but a=7

49-d²-7=6

=-d²-42=6

=-d²=-36

=d²=36

=d=6 or d=-6

So

AP=7+6,7,7-6

=13,7,1.

AP can also be 7-6,7,7+6

=1,7,13

Hope it helps.