Question

The sum of the first three terms of an A.P.is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.

Answer 1

Let the three numbers in A.P. are

a – d, a, a + d

Now, a – d + a + a + d = 33

⇒ 3a = 33

⇒ a = `(33)/(3)` = 11

And (a – d)(a + d) = a + 29

a² – d² = a + 29

(11)² – d² = 11 + 29

⇒ 121 – d² = 40

d² = 121 – 40

= 81

= (±9)²

∴ d = ±9

If d = 9, then

∴ Numbers are 11 –9, 11, 11 + 9

⇒ 2, 11, 20

If d = –9, then

11 + 9, 11, 11 – 9

⇒ 20, 11, 2

Hence numbers are 2, 11, 20 or 20, 11, 2.