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If the sum of three terms of G.P. is 19 and product is 216 , then the common ratio of the series is
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Verified Answer
The correct answer is:
$\frac{3}{2}$
Let three terms of G.P. are $a, a r, a r^2$. Then
$a+a r+a r^2=19 \Rightarrow a\left[1+r+r^2\right]=19$\ldots(i)
a. $a r \cdot a r^2=216 \Rightarrow a^3 r^3=216 \Rightarrow a r=6$
\ldots(ii)
Dividing (ii) by (i),
$\frac{6}{r}+\frac{6}{r} r+\frac{6}{r} r^2=19 \Rightarrow \frac{6}{r}+6+6 r=19$
$\Rightarrow r^2-\frac{13}{6} r+1=0$ Hence $\quad r=\frac{3}{2}$
$a+a r+a r^2=19 \Rightarrow a\left[1+r+r^2\right]=19$\ldots(i)
a. $a r \cdot a r^2=216 \Rightarrow a^3 r^3=216 \Rightarrow a r=6$
\ldots(ii)
Dividing (ii) by (i),
$\frac{6}{r}+\frac{6}{r} r+\frac{6}{r} r^2=19 \Rightarrow \frac{6}{r}+6+6 r=19$
$\Rightarrow r^2-\frac{13}{6} r+1=0$ Hence $\quad r=\frac{3}{2}$
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