Question

Let $a_n$ be the $n^{th}$ term of a G.P. of positive terms. If $\sum _{n=1}^{100}{a}_{2n+1}=200$ and $\sum _{n=1}^{100}{a}_{2n}=100$ then value of $\begin{array}{l}\sum _{n=1}^{200}{a}_n\end{array}$ is

• A. $300$
• B. $175$
• C. $225$
• D. $150$

Answer 1

The correct option is D

$150$

Explanation for the correct option:

Finding the value of $a_n$:

Considering $a_n$ is a positive term of GP

Let GP be $a,ar,a{r}^2,\dots .$

Given that $\begin{array}{l}\sum _{n=1}^{100}{a}_{2n+1}=200\end{array}$

$$\begin{gathered} \Rightarrow a_3+a_5+a_7+\dots +a_{201}=200 \\ \Rightarrow a{r}^2+a{r}^4+\dots +a{r}^{200}=200 \end{gathered}$$

Since, we know that sum of $n^{th}$ term of GP with first element $a$ and common ratio $r$

$\Rightarrow \frac{a\left(r^n-1\right)}{r-1};r>1$

$\therefore \begin{array}{l}\frac{a{r}^2(r^{200}-1)}{r^2-1}\end{array}=200....\left(i\right)$

Similarly,

$$\begin{gathered} \Rightarrow \begin{array}{l}\sum _{n=1}^{100}{a}_{2n}=100\end{array} \\ \Rightarrow a_2+a_4+\dots +a_{200}=100 \\ \Rightarrow ar+a{r}^3+\dots a{r}^{199}=100 \\ \Rightarrow \begin{array}{l}\frac{ar(r^{200}-1)}{r^2-1}\end{array}=100...\left(ii\right) \end{gathered}$$

By dividing equations $\left(2\right)\text{by}\left(1\right)$ we get

$$\begin{gathered} \therefore \begin{array}{l}\frac{\frac{a{r}^2(r^{200}-1)}{r^2-1}}{\frac{ar(r^{200}-1)}{r^2-1}}\end{array}=\frac{200}{100} \\ \Rightarrow r=2 \end{gathered}$$

Adding equations $\left(i\right)\&\left(ii\right)$

$$\begin{gathered} \Rightarrow \begin{array}{l}\sum _{n=1}^{100}{a}_{2n+1}+\sum _{n=1}^{100}{a}_{2n}=200\end{array}+100 \\ \Rightarrow a_2+a_3+a_4\dots .+a_{200}+a_{201}=300 \\ \Rightarrow ar+a{r}^2+a{r}^3+\dots a{r}^{200}=300 \\ \Rightarrow r(a+ar+a{r}^2+\dots +a{r}^{199})=300\left[\text{substituting}r=2\right] \\ \Rightarrow 2(a_1+a_2+a_3+\dots +a_{200})=300 \\ \Rightarrow \begin{array}{l}\sum _{n=1}^{200}{a}_n=150\end{array} \end{gathered}$$

Hence, the correct answer is option (D).