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If the sum of the first two terms and the sum of the first four terms of a geometric progression with positive common ratio are 8 and 80 respectively, then what is the 6 th term?
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Verified Answer
The correct answer is:
486
Let the Geometric progression be a, ar, ar $^{2}$, ar $^{3}, \ldots$ with common ratio $\mathrm{r}$ and first term 'a '. According to the question, we have
$$
\mathrm{a}+\mathrm{ar}=8 \Rightarrow \mathrm{a}(1+\mathrm{r})=8 ...(i)
$$
and $\mathrm{a}+\mathrm{ar}+\mathrm{ar}^{2}+\mathrm{ar}^{3}=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})+\operatorname{ar}^{2}(1+\mathrm{r})=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})\left(1+\mathrm{r}^{2}\right)=80$
$\Rightarrow 8\left(1+r^{2}\right)=80 \quad$ (from (i))
$\Rightarrow 1+r^{2}=\frac{80}{8}=10$
$\Rightarrow \mathrm{r}^{2}=10-1=9$
$\Rightarrow \mathrm{r}=3 \quad(\because \mathrm{r}>0)$
From eq. (i), $a(1+3)=8$ $\Rightarrow \mathrm{a}=2$
Now, $6^{\text {th }}$ term $=$ ar $^{5}=2(3)^{5}=2 \times 243=486$
$$
\mathrm{a}+\mathrm{ar}=8 \Rightarrow \mathrm{a}(1+\mathrm{r})=8 ...(i)
$$
and $\mathrm{a}+\mathrm{ar}+\mathrm{ar}^{2}+\mathrm{ar}^{3}=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})+\operatorname{ar}^{2}(1+\mathrm{r})=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})\left(1+\mathrm{r}^{2}\right)=80$
$\Rightarrow 8\left(1+r^{2}\right)=80 \quad$ (from (i))
$\Rightarrow 1+r^{2}=\frac{80}{8}=10$
$\Rightarrow \mathrm{r}^{2}=10-1=9$
$\Rightarrow \mathrm{r}=3 \quad(\because \mathrm{r}>0)$
From eq. (i), $a(1+3)=8$ $\Rightarrow \mathrm{a}=2$
Now, $6^{\text {th }}$ term $=$ ar $^{5}=2(3)^{5}=2 \times 243=486$
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