Question

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds these term by 6, find three terms.

Answer 1

The three terms can be 1,7,13 or 13,7,1.

Let the common difference in AP be 'd'.

Thus, the terms in AP are a, a+d, a+2d.

Given that the sum of the 3 numbers equal to 21.

So, (a+a+d+a+2d) = 21

⇒ 3a + 3d = 21

⇒ a+d = 21/3 = 7

Product of the first and third terms exceed the second by 6.

So, a(a+2d) - (a+d) = 6

⇒ a² + 2ad - 7 = 6                             [a+d = 7]

⇒ a² + 2a(7-a) - 13 = 0                      [d = 7-a]

⇒ a² + 14a - 2a² - 13 = 0

⇒ 14a -a² - 13 = 0

⇒ a² -14a + 13 = 0

⇒ a² -a -13a +13 = 0

⇒ a(a-1) -13(a-1) = 0

⇒ (a-1)(a-13) = 0

By zero product rule, a = 1, 13

For a = 1,

d = 7-a = 7-1 = 6

So, a, a+d, a+2d are  1, (1+6), (1+12)

=1,7,13

When a = 13,

d = 7-a = 7-13 = -6

So, a, a+d, a+2d are  13, (13-6), (13-12)

=13,7,1