Find the 10th and nth terms of the GP

Question:

Find the 10th and nth terms of the GP $-\frac{3}{4}, \frac{1}{2},-\frac{1}{3}, \frac{2}{9} \ldots$

 

Solution:

Given GP is $-\frac{3}{4}, \frac{1}{2},-\frac{1}{3}, \frac{2}{9} \ldots .$

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

Where $r$ is the common ratio.

The first term in the given GP, $\mathrm{a}=\mathrm{a}_{1}=-\frac{3}{4}$

The second term in $\mathrm{GP}, \mathrm{a}_{2}=\frac{1}{2}$

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

$r=-\frac{\frac{1}{2}}{\frac{3}{4}}=-\frac{2}{3}$

Now, $n^{\text {th }}$ term of GP is, $a_{n}=a r^{n-1}$

So, the $10^{\text {th }}$ term. $a_{10}=a r^{9}$

$a_{10}=a r^{9}=\left(-\frac{3}{4}\right)\left(-\frac{2}{3}\right)^{9}=\frac{128}{6561}$

Now, the required $\mathrm{n}^{\text {th }}$ term, $\mathrm{an}=\mathrm{ar}^{\mathrm{n}-1}$

$a_{n}=\left(-\frac{3}{4}\right)\left(-\frac{2}{3}\right)^{n-1}=\left(\frac{9}{8}\right)\left(-\frac{2}{3}\right)^{n}$

Hence, the $10^{\text {th }}$ term, $\mathrm{a}_{10}=\frac{128}{6561}$ and $\mathrm{n}^{\text {th }}$ term,

$a_{n}=\left(\frac{9}{8}\right)\left(-\frac{2}{3}\right)^{n}$

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