Find the number of terms of the given series: $2 - 2\sqrt{2} + 4 - 4\sqrt{2} + \dots + 16$.

Question

In this question, common ratio will be $-\sqrt{2}$ and if I use this value ($-\sqrt{2}$) to find the number of terms, I cannot get the answer. Before, I solved using "$-\sqrt{2} = \sqrt{2}$" which is wrong method. Can you please suggest me to solve this?

Answer 1

I'm upgrading my comment to an answer, since it seems to have settled matters for OP.

We're trying to solve $(-\sqrt2)^{n-1}=8$. If $n$ is even, then $n-1$ is odd, and then $(-\sqrt2)^{n-1}$ is negative, and can't equal $8$. So if $n$ exists, it has to be odd, and then $(-\sqrt2)^{n-1}$ is the same thing as $(\sqrt2)^{n-1}$, and that's why the minus sign isn't there any more. Of course, we have to check when we're done solving that $n$ really is odd, and since it turns out $n=7$, we're OK.

Answer 2

In GP $T_n=a r^{n-1}$ here $2$ and $r=-\sqrt{2}$, then $$2 (-\sqrt{2})^{n-1}=16 \implies (-\sqrt{2})^{n-1}=8=2^3 \implies n=7$$

Answer 3

Since the $n$th term is $2(-\sqrt{2})^{n-1}$ then we have

$$2(-\sqrt{2})^{n-1}=16 \implies (-\sqrt{2})^{n-1}=8=2^{3}=(\sqrt{2})^{6}=(-\sqrt{2})^{6}$$

and hence $n=7$. Here we have used the facts that $\sqrt{a}\sqrt{a}=a$ and $(-1)^{k}=1$ for even $k$.