Question

The sum of three consecutive terms of an ap is 6 and their product is - 120 find the three terms let three numbers be a-d,a
a + b

Answer 1

Answer:-

The three terms in AP is 2, -6 , 10.

Given:-

Sum of three consecutive terms in A. P is 6.

Their product is -120.

To find :-

The three terms.

Solution:-

Let the three terms in A. P be a

a + d, a - d.

A/Q

→$a + a+d + a-d = 6$

→$3a = 6$

→$a = \dfrac{6}{3}$

→$a = 2$

→$a (a+d) (a-d) = -120$

→$a ( a^2 -d^2) = -120$

• Put the value of a

→$2 ( 4 -d^2 ) = -120$

→$8 - 2d^2 = -120$

→$8 +120 = 2d^2$

→$128 = 2 d^2$

→$d^2 = \dfrac{128}{2}$

→$d^2 = 64$

→$d= \sqrt{64}$

→$d = \pm 8$

The three terms in A. P will be :-

→a = 2

→a - d = 2 - 8 = -6

→a + d = 2 + 8 = 10

• Taking d = -8

→a = 2

→a + d = 2 - 8 = -6

→a - d = 2 -(-8) = 2 + 8 = 10

hence,

The three terms in AP is 2, -6 , 10.