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If a, b, care in geometric progression and a, $2 \mathrm{~b}, 3 \mathrm{c}$ are in arithmetic progression, then what is the common ratio $\mathrm{r}$ such that $0 < \mathrm{r} < 1 ?$
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The correct answer is:
$\frac{1}{3}$
Given that a, b, c, are in GP. Let r be common ratio of GP. So, $a=a, b=a r$ and $c=a r^{2}$ ...(i)
Also, given that $a, 2 b, 3 c$ are in AP.
$\Rightarrow 2 b=\frac{a+3 c}{2}$
$\Rightarrow \quad 4 \mathrm{~b}=\mathrm{a}+3 \mathrm{c}$
From Eq. (1) $4 a r=a+3 a r^{2}$
$\Rightarrow \quad 3 \mathrm{ar}^{2}-4 \mathrm{ar}+\mathrm{a}=0$
$\Rightarrow \quad 3 r^{2}-4 r+1=0$
$\Rightarrow \quad 3 r^{2}-3 r-r+1=0$
$\Rightarrow 3 \mathrm{r}(\mathrm{r}-1)-1(\mathrm{r}-1)=0$
$\Rightarrow(\mathrm{r}-1)(3 \mathrm{r}-1)=0$
$\Rightarrow \mathrm{r}=1$ or $\mathrm{r}=\frac{1}{3}, \mathrm{r}=\frac{1}{3}$ is in the choice.
Also, given that $a, 2 b, 3 c$ are in AP.
$\Rightarrow 2 b=\frac{a+3 c}{2}$
$\Rightarrow \quad 4 \mathrm{~b}=\mathrm{a}+3 \mathrm{c}$
From Eq. (1) $4 a r=a+3 a r^{2}$
$\Rightarrow \quad 3 \mathrm{ar}^{2}-4 \mathrm{ar}+\mathrm{a}=0$
$\Rightarrow \quad 3 r^{2}-4 r+1=0$
$\Rightarrow \quad 3 r^{2}-3 r-r+1=0$
$\Rightarrow 3 \mathrm{r}(\mathrm{r}-1)-1(\mathrm{r}-1)=0$
$\Rightarrow(\mathrm{r}-1)(3 \mathrm{r}-1)=0$
$\Rightarrow \mathrm{r}=1$ or $\mathrm{r}=\frac{1}{3}, \mathrm{r}=\frac{1}{3}$ is in the choice.
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