If {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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8. Sequences and Series

easy

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

A

$6$

B

$36$

C

$\frac{4}{9}$

D

$\frac{9}{4}$

Solution

(a) Accordingly, $a{r^9} = 9$ and $a{r^3} = 4$

$ \Rightarrow $ ${r^3} = \frac{3}{2}$ and $a = \frac{8}{3}$.

$\therefore $ ${7^{th}}$ term $i.e.$ $a{r^6} = \frac{8}{3}{\left( {\frac{3}{2}} \right)^2} = 6$.

Trick : ${7^{th}}$ term is equidistant from ${10^{th}}$ and ${4^{th}}$ so it will be $\sqrt {9 \times 4} = 6$.

Standard 11

Mathematics

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