If the {10^{th}} term of a geometric progress | vedclass

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Question

(a) Accordingly, $a{r^9} = 9$ and $a{r^3} = 4$

$ \Rightarrow $ ${r^3} = \frac{3}{2}$ and $a = \frac{8}{3}$.

$	herefore $ ${7^{th}}$ term $i.e.$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(a) Accordingly, $a{r^9} = 9$ and $a{r^3} = 4$

$ \Rightarrow $ ${r^3} = \frac{3}{2}$ and $a = \frac{8}{3}$.

$	herefore $ ${7^{th}}$ term $i.e.$ $a{r^6} = \frac{8}{3}{\left( {\frac{3}{2}} ight)^2} = 6$.

Trick : ${7^{th}}$ term is equidistant from ${10^{th}}$ and ${4^{th}}$ so it will be $\sqrt {9 	imes 4} = 6$.

If the ${10^{th}}$ term of a geometric progression is $9$ and ${4^{th}}$ term is $4$, then its ${7^{th}}$ term is

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