If the sum of three terms of G.P. is 19 and p | vedclass

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Question

(b) Let three terms of $G.P.$ are $a,\;ar,\;a{r^2}$. Then

$a + ar + a{r^2} = 19 $

$\Rightarrow a[1 + r + {r^2}] = 19$ &hellip;..$(i)$

$a\;.\;

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(b) Let three terms of $G.P.$ are $a,\;ar,\;a{r^2}$. Then

$a + ar + a{r^2} = 19 $

$\Rightarrow a[1 + r + {r^2}] = 19$ &hellip;..$(i)$

$a\;.\;ar\;.\;a{r^2} = 216$

$\Rightarrow {a^3}{r^3} = 216$

$\Rightarrow ar = 6$ &hellip;..$(ii)$

Dividing $(ii)$ by $(i),$

$\frac{6}{r} + \frac{6}{r}r + \frac{6}{r}{r^2} = 19 $

$\Rightarrow \frac{6}{r} + 6 + 6r = 19$

$ \Rightarrow {r^2} - \frac{{13}}{6}r + 1 = 0$.

Hence $r = \frac{3}{2}$.

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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