Question

Find three consecutive terms in A.P. whose sum is 9 and the product of their cubes is 3375

Answer 1

Let the 3 consecutive terms in A.P. be,

(a-d), a, (a+d)

According to given condition,

a-d+a+a-d = 9

3a-2d = 9 ==> a = (9+2d)/3

Also,

(a-d)³a³(a+d)³ = 3375

(a³-3a²d+3ad²-d³)a³(a³+3a²d+3ad²+d³) = 3375

By solving this, we get the three consecutive terms which are in A.P.

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