Question

The sum of first three terms of an AP is 36 and the product of the first and third terms exceed the second term by 51, find the AP.

Answer 1

Let

\(a_1 = a -d\)

\(a_2 = a\)

\(a_3 =a +d\)

\(\therefore \) sum \(=36\)

⇒ \(3a = 36\)

⇒ \(a =12\)

Also,

\(a^2 - d^2 = a + 51\)

\(= 12 + 51\)

\(= 63\)

⇒ \(12^2 - d^2 = 63\)

⇒ \(d^2 = 144 - 63 = 81\)

⇒ \(d = \pm 9\)

\(\therefore \) Required terms are 3, 12, 21.