Let a_n be the nth term of a G.P.

Question

Let $a_n$ be the $n$th term of a G.P. of positive terms. If $\displaystyle\sum_{n=1}^{100}a_{2n+1}=200$ and $\displaystyle\sum_{n=1}^{100}a_{2n}=100$, then $\displaystyle\sum_{n=1}^{200}a_n$ is equal to:

Accepted Answer

$\displaystyle\sum_{n=1}^{100}a_{2n+1} = a_3+a_5+\cdots+a_{201} = 200$, and $\displaystyle\sum_{n=1}^{100}a_{2n} = a_2+a_4+\cdots+a_{200}=100$.

Dividing (each term of the first sum is $r$ times the corresponding term of the second, shifted): $r=2$.

$$a_2+a_3+\cdots+a_{201} = 300 \implies r\sum_{n=1}^{200}a_n = 300 \implies \sum_{n=1}^{200}a_n = \frac{300}{2}=150.$$

Solutions · 1