If the $n^{t h}, (2 n)^{t h}, (3 n)^{t h}$ terms of a G.P. are a, b, c respectively then

b^{2} = a c .

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a^{2} = b c .

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c^{2} = b a

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None of these

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Solution

The correct option is A $b^{2} = a c .$
Let A an R be the first term and common ratio of given GP respectively. Then,
$n^{t h} t e r m = a \Rightarrow A R^{(n - 1)} = a$ ...(1)
${2 n}^{t h} t e r m = 6 \Rightarrow A R^{(2 n - 1)} = b$ ...(2)
${3 n}^{t h} t e r m = a \Rightarrow A R^{(3 n - 1)} = c$ ...(3)
Multiplying (1) and (3), we get
$A^{2} R^{4 n - 2} = a c$
$\Rightarrow (A R^{2 n - 1})^{2} = a c$
$\Rightarrow b^{2} = a c$
Hence proved.

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