### Sequences And Series Question 2

#### Question 2 - 24 January - Shift 1

The $4^{\text{th }}$ term of GP is 500 and its common ratio is $\frac{1}{m}, m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this GP. If $S_6>S_5+1$ and $S_7<S_6+\frac{1}{2}$, then the number of possible values of $m$ is _______

## Show Answer

#### Answer: 12

#### Solution:

#### Formula: Geometric Progression ($ n^{th}$ term ) and (Sum of $n^{th}$ term)

$T_4=500 \quad$ where $a=$ first term,

$ r=\text{ common ratio }=\frac{1}{m}, m \in N $

$ar^{3}=500$

$\frac{a}{m^{3}}=500$

$S_n-S _{n-1}=ar^{n-1}$

$S_6>S_5+1 \quad$ and $S_7-S_6<\frac{1}{2}$

$S_6-S_5>1$

$\frac{a}{m^{6}}<\frac{1}{2}$

$ar^{5}>1$

$m^{3}>10^{3}$

$\frac{500}{m^{2}}>1$

$m>10$

$m^{2}<500$

From (1) and (2)

$m=11,12,13$.

So number of possible values of $m$ is 12