Find three consecutive terms in an A.P. whose sum is – 3 and the product of their cubes is 512.

3. Find three consecutive terms in an A.P. whose sum is – 3 and the product of their cubes is 512.

Sol. Let three consecutive terms in an A.P. be a – d, a, a + d

As per the first given condition,

a – d + a + a + d = – 3

∴  3a = – 3

∴  a = – 1 ............. eq. (i)

As per the second given condition,

(a – d)³ a³ (a + d)³ = 512

∴  [(a – d) a (a + d)]³ = 512

Taking cube roots on both sides, we get

(a – d) a (a + d) = 8

∴  a (a – d) (a + d) = 8

∴  a (a² – d²) = 8  

[∵ a² – b² = (a + b)(a – b)]

∴  – 1 [(– 1)² – d²] = 8 [from eq. (i)]

∴  – 1 (1 – d²) = 8

∴  d² – 1 = 8

∴  d² = 8 + 1

∴  d² = 9

∴  d = ± 3

If a = -1 and d = 3 then	If a = -1 and d = -3 then
a – d = - 1 – 3 = - 4	a – d = - 1 – (-3)= -1 + 3 = 2
a = - 1	a = - 1
a + d = -1 + 3 = 2	a + d = -1 +(-3) = -1 – 3= -4

The three consecutive terms of A.P. are – 4, – 1, 2 or 2, –1, –4