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# Find the 17th and nth terms of the

Question:

Find the $17^{\text {th }}$ and nth terms of the GP $2,2 \sqrt{2}, 4,8 \sqrt{2} \ldots .$

Solution:

Given GP is $2,2 \sqrt{2}, 4,8 \sqrt{2} \ldots \ldots$

The given GP is of the form, $a, a r, a r^{2}, a r^{3} \ldots .$

Where $r$ is the common ratio.

First term in the given GP, $a_{1}=a=2$

Second term in GP, $a_{2}=2 \sqrt{2}$

Now, the common ratio, $\mathrm{r}=\frac{\mathrm{a}_{2}}{\mathrm{a}_{1}}$

$r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$

Now, $\mathrm{n}^{\text {th }}$ term of GP is, $\mathrm{a}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$

So, the $17^{\text {th }}$ term in the GP,

$a_{17}=a r^{16}$

$=2 \times(\sqrt{2})^{16}$

$=512$

$\mathrm{n}^{\text {th }}$ term in the GP,

$a_{n}=a r^{n-1}$

$=2(\sqrt{2})^{n-1}$

$=(\sqrt{2})^{n+1}$

Hence, $17^{\text {th }}$ term $=512$ and $\mathrm{n}^{\text {th }}$ term $=(\sqrt{2})^{\mathrm{n}+1}$