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
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If a = -1 and d = -3 then
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a – d = - 1 – 3 = - 4
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a – d = - 1 – (-3)= -1 + 3 = 2
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a = - 1
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a = - 1
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a + d = -1 + 3 = 2
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a + d = -1 +(-3) = -1 – 3= -4
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The three consecutive terms of A.P. are – 4, – 1, 2 or 2, –1, –4
