JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 20)
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 :
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$ and $$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$,
then $$\sum\limits_{n = 1}^{200} {{a_n}} $$ is equal to :
150
175
225
300
Explanation
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$
$$ \Rightarrow $$ 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
$$ \Rightarrow $$ 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
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