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

Question

Let an be the nth term of a G.P. of positive terms.

$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$ and $$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$,

then $$\sum\limits_{n = 1}^{200} {{a_n}} $$ is equal to :

Accepted Answer

$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$

$$ \Rightarrow $$ a3 + a5 + a7 + .... + a201 = 200

$$ \Rightarrow $$ $$a{r^2}{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 200 ....(1)

$$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$

$$ \Rightarrow $$ a2 + a4 + a6 + ... + a200 = 100

$$ \Rightarrow $$ $$ar{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 100  ....(2)

dividing (1) by (2)

we get, r = 2

adding both  (1) and (2), we get

a2 + a3 + a4 + a5 + ..... + a201 = 300

$$ \Rightarrow $$ r(a1 + a2 + ..... + a200) = 300

$$ \Rightarrow $$ a1 + a2 + ..... + a200 = $${{300} \over r}$$

$$ \Rightarrow $$ $$\sum\limits_{n = 1}^{200} {{a_n}} $$ = $${{300} \over 2}$$ = 150

Solutions · 1