Which term of the GP
Question:

Which term of the GP $\frac{1}{4}, \frac{-1}{2}, 1 \ldots . .$ is $-128 ?$

Solution:

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

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, $\mathrm{a}=\mathrm{a}_{1}=\frac{1}{4}$

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

Now, the common ratio, $r=\frac{a_{2}}{a_{1}}$

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

Let us consider $-128$ as the $\mathrm{n}^{\text {th }}$ term of the GP.

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

$-128=\left(\frac{1}{4}\right)(-2)_{n-1}$

$(-2)^{n}=1024=(-2)^{10}$

$\mathrm{n}=10$

So, $-128$ is the $10^{\text {th }}$ term in GP.