InterviewSolution
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InterviewSolution
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Does there exist a geometric progression containing 27,8 and 12 as three of its term ? If it exists, then how many such progressions are possible ? |
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Answer» Correct Answer - yes, infinite Let 27, 8, 12 be three terms of a GP `rArr t_(m) = 27, t_(n) = 8 and t_(p) = 12` `AR^(m-1) = 27, AR^(n-1) = 8` and `AR^(p -1) = 12` `:. R = ((27)/(8))^(1//(m-n)) and R = ((8)/(12))^(1//(n -p))` `rArr ((27)/(8))^(1//(m-n)) = ((2)/(3))^(1//(n -p))` `rArr 3^(3//(m-n)).3^(1//(n -p)) = 2^(1//(n-p)).2^(3//(m-n))` `rArr (3^((3)/(m-n) + (1)/(n-p)))/(2^((1)/(n-p) + (3)/(m-n))) = 1` `:. (3)/(m-n) + (1)/(n-p) = 0 and (1)/(n-p) + (3)/(m -n) = 0` `rArr 3(n -p) = n - m and 2n = 3p - m` Hence, there exists infinite GP for which 27, 8 and 12 as three of its terms. |
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