"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 :"

Question

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"

Let a_n be the n^th 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 :

Solution

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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 :

Solution

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Let a_n be the n^th 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 :