Question

three consecutive terms in an A.P whose sum is -

3 and the product of their cubes 512

Answer 1

Hey buddy here is ur answer .....


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 .

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Answer 2

Answer:

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