If the 6th, 11th and 16th terms of a GP are a,b,c respectively then

a + c = 2 b

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b^{2} = a c

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a^{2} + c^{2} = b^{2}

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none of these

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Solution

The correct option is D $b^{2} = a c$
Let the first term of GP be $A$ and common ratio $r$
Given
$a = A r^{5}, b = A r^{10}, c = A r^{15}$
Clearly, $(A r^{10})^{2} = A r^{5} . A r^{15}$
$∴ a, b, c$ are in GP.

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