If the 4th, 7th and 10th terms of a GP are a, b and c respectively, then the relation between a, b and c is

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$a^{2} = b c$

$c^{2} = a b$

$b = \frac{a + c}{2}$

$b^{2} = a c$

answer is C.

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Detailed Solution

Given $\begin{aligned} a_{4} = a \\ a_{7} = b \\ a_{10} = c \end{aligned}$

Let A be the first term and R be the common ratio of GP, then

$\begin{aligned} A R^{3} = a \\ A R^{6} = b \\ A R^{9} = c \end{aligned}$

On dividing Eq. (v) by Eq. (iv), we get

$\begin{aligned} R^{3} = \frac{b}{a} \\ R = {(\frac{b}{a})}^{1 / 3} \end{aligned}$

Again, dividing Eq. (vi) by Eq. (v), we get

$\begin{aligned} R^{3} = \frac{c}{b} \\ R = {(\frac{c}{b})}^{1 / 3} \end{aligned}$

From Eqs. (vii) and (viii). ${(\frac{b}{a})}^{1 / 3} = {(\frac{c}{b})}^{1 / 3}$

On cubing both sides, we get

$\begin{aligned} \frac{b}{a} = \frac{c}{b} \\ b^{2} = a c \end{aligned}$

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