JEE Main 2023 (Online) 24th January Morning ShiftQuestion 1
The $4^{th}$ term of a GP is $500$ and its common ratio is $\frac1m$, $m\in\mathbb N$. Let $S_n$ denote the sum of the first $n$ terms of this GP. If $S_6>S_5+1$ and $S_7<S_6+\frac12$, then the number of possible values of $m$ is ______
Sequence and Series · Geometric Progression
Answer$12$
Solution
Formulas used. (1) $S_n-S_{n-1}=a_n$(2) $a_n=a_4\,r^{n-4}$Step 1 - The terms. With $r=\dfrac1m$: $$a_6=\frac{500}{m^2},\qquad a_7=\frac{500}{m^3}.$$ Step 2 - First condition. $$a_6>1\ \Longrightarrow\ m^2<500\ \Longrightarrow\ m\le22 .$$ Step 3 - Second condition. $$a_7<\frac12\ \Longrightarrow\ m^3>1000\ \Longrightarrow\ m\ge11 .$$ Step 4 - Count. $m=11,12,\dots,22$: $12$ values. Answer: 12