JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
Let $a_n$ be the $n^{\text{th}}$ 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 $\sum_{n=1}^{200}a_n$ is equal to :
Sequence and Series · Geometric Progression
Solution
Formulas used. (1) In a G.P., $a_{2n+1}=r\,a_{2n}$ and $a_{2n-1}=\frac{a_{2n}}{r}$Step 1 - Ratio. $a_3+a_5+\cdots+a_{201}=r\left(a_2+a_4+\cdots+a_{200}\right)$: $$200=100r\ \Longrightarrow\ r=2 .$$ Step 2 - Odd terms up to $a_{199}$. $$a_1+a_3+\cdots+a_{199}=\frac1r\left(a_2+\cdots+a_{200}\right)=\frac{100}{2}=50 .$$ Step 3 - Total. $$\sum_{n=1}^{200}a_n=50+100=150 .$$ Answer: (B)