Let a_{n} be the n^{text {th }} term of a G.P | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\sum_{n=1}^{100} a_{2 n+1}=200 \Rightarrow a_{3}+a_{5}+a_{7}+\ldots .+a_{201}=200$

$\Rightarrow \operatorname{ar}^{2} \frac{\left(\mathrm{r}^{200}

Vedclass · Accepted Answer

$150$

Vedclass · Answer

$\sum_{n=1}^{100} a_{2 n+1}=200 \Rightarrow a_{3}+a_{5}+a_{7}+\ldots .+a_{201}=200$

$\Rightarrow \operatorname{ar}^{2} \frac{\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=200$

$\sum_{n=1}^{100} a_{2 n}=100 \Rightarrow a_{2}+a_{4}+a_{6}+\ldots+a_{200}=100$

$\Rightarrow \frac{\operatorname{ar}\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=100$

On dividing $\mathrm{r}=2$

on adding $a_{2}+a_{3}+a_{4}+a_{5}+\ldots+a_{200}+a_{201}=300$

$\Rightarrow \mathrm{r}\left(\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{200}\right)=300$

$\Rightarrow \sum_{n=1}^{200} a_{n}=150$

Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms.

If $\sum\limits_{n=1}^{100} a_{2 n+1}=200$ and $\sum\limits_{n=1}^{100} a_{2 n}=100,$ then $\sum\limits_{n=1}^{200} a_{n}$ is equal to

Similar Questions

If $a,\;b,\;c$ are in $A.P.$, $b,\;c,\;d$ are in $G.P.$ and $c,\;d,\;e$ are in $H.P.$, then $a,\;c,\;e$ are in

The value of $0.\mathop {234}\limits^{\,\,\, \bullet \,\, \bullet } $ is

If the first term of a $G.P.$ ${a_1},\;{a_2},\;{a_3},..........$ is unity such that $4{a_2} + 5{a_3}$ is least, then the common ratio of $G.P.$ is

The $G.M.$ of roots of the equation ${x^2} - 18x + 9 = 0$ is

How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$

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