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Let $a, b$ and $c$ be in G.P. with common ratio $r$ where $a \neq 0 and 0 < r \leq 1 / 2$ . If $3 a, 7 b$ and $15 c$ are the first three terms of an A.P., then the 4th term of this A.P. is:

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detailed solution

Correct option is B

Given, $a, b$ and $c$ are in G.P. with common ratio $r$ .

$∴ b = a r$ and $c = a r^{2}$ ……………… …(i)

Also, $3 a, 7 b$ and $15 c$ are in A.P. [Given]

$\begin{array}{ll} ∴ 3 a + 15 c = 14 b \end{array}$

$\Rightarrow 3 a + 15 a r^{2} = 14 a r$ [Using (i)]

$\begin{array}{l} \Rightarrow 15 r^{2} - 14 r + 3 = 0 \Rightarrow 15 r^{2} - 9 r - 5 r + 3 = 0 \\ \Rightarrow (3 r - 1) (5 r - 3) = 0 \end{array}$

$\Rightarrow r = 1 / 3$ or $3 / 5 \Rightarrow r = 1 / 3$ $[∵ r \in (0, 1 / 2]]$

$∴$ The required A.P. is $3 a, 7 a / 3, 5 a / 3, \dots$

Hence, 4th term is $3 a / 3 = a$

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Let a,b and c be in G.P. with common ratio r where a≠0 and 0<r≤1/2. If 3a, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is: