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If the 10 th term of a GP is 9 and 4 th term is 4 , then what is its 7th term?
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Verified Answer
The correct answer is:
6
General term of G.P= ar $^{\text {1-1 }}$. Given : $\mathrm{a}_{10}=9$ and $\mathrm{a}_{4}=4$
$\Rightarrow \mathrm{ar}^{9}=9$ and $\mathrm{ar}^{3}=4$
On dividing we get
$\begin{aligned} \frac{a r^{9}}{a r^{3}}=& \frac{9}{4} \Rightarrow r^{6}=\frac{9}{4} \\ a r^{3}=4 & \Rightarrow\left(a r^{3}\right)^{2}=16 \\ & \Rightarrow a^{2} r^{6}=16 \\ & \Rightarrow a^{2} \times \frac{9}{4}=16 \Rightarrow a^{2}=\frac{64}{9} \Rightarrow a=\frac{8}{3} \\ \text { Thus, } a_{7}=a r^{6}=\frac{8}{3} \times \frac{9}{4}=6 \end{aligned}$
$\Rightarrow \mathrm{ar}^{9}=9$ and $\mathrm{ar}^{3}=4$
On dividing we get
$\begin{aligned} \frac{a r^{9}}{a r^{3}}=& \frac{9}{4} \Rightarrow r^{6}=\frac{9}{4} \\ a r^{3}=4 & \Rightarrow\left(a r^{3}\right)^{2}=16 \\ & \Rightarrow a^{2} r^{6}=16 \\ & \Rightarrow a^{2} \times \frac{9}{4}=16 \Rightarrow a^{2}=\frac{64}{9} \Rightarrow a=\frac{8}{3} \\ \text { Thus, } a_{7}=a r^{6}=\frac{8}{3} \times \frac{9}{4}=6 \end{aligned}$
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