Which term of the GP


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



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, $r=\frac{a_{2}}{a_{1}}$


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

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




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


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