Question

find 3 consecutive term in an AP whose sum is 3 and product of their cube is 512​

Answer 1

Answer:

Let the 3 terms in AP be

a-d, a, a+d

Given

a-d+a+a+d = 3

3a = 3 - - - - -> a = 1

Also

[(a-d)(a)(a+d)]^3= 512

[(1-d)(1)(1+d)]^3 = 8^3

Cube rooting on both the sides

(1-d)(1)(1+d) = 8

1-d^2 = 8

d^2 = -7

NO SOLUTION.

PLEASE CHECK THE QUESTION