Let a, b and c be in G.P with common ratio r, | vedclass

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Question

$b=ar$

$c = a{r^2}$

$3a.7b$ and $15c$ are in$A.P,$

$ \Rightarrow 14b = 3a + 15c$

$ \Rightarrow 14\left( {ar} \right) = 3a + 15a{r^2}$

$

Vedclass · Accepted Answer

$a$

Vedclass · Answer

$b=ar$

$c = a{r^2}$

$3a.7b$ and $15c$ are in$A.P,$

$ \Rightarrow 14b = 3a + 15c$

$ \Rightarrow 14\left( {ar} ight) = 3a + 15a{r^2}$

$ \Rightarrow 14r = 3 + 15{r^2}$

$ \Rightarrow 15{r^2} - 14r + 3 = 0\,\,\,\,\,\, \Rightarrow \left( {3r - 1} ight)\left( {5r - 1} ight) = 0$

$r = \frac{1}{3},\frac{3}{5}$

Only acceptable value is $r = \frac{1}{3}$, because $r \in \left( {0,\frac{1}{2}} ight)$

$	herefore c.d = 7b - 3a = 7ar - 3a = \frac{7}{3}a - 3a =  - \frac{2}{3}a$

$	herefore {4^{th}}$ term $ = 15c - \frac{2}{3}a = \frac{{15}}{{9a}} - \frac{2}{3}a = a$

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

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