JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 16)
The 4$$^\mathrm{th}$$ term of GP is 500 and its common ratio is $$\frac{1}{m},m\in\mathbb{N}$$. Let $$\mathrm{S_n}$$ denote the sum of the first n terms of this GP. If $$\mathrm{S_6 > S_5 + 1}$$ and $$\mathrm{S_7 < S_6 + \frac{1}{2}}$$, then the number of possible values of m is ___________
Answer
12
Explanation
$T_{4}=500$
$$ a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}} $$
Now,
$$ \begin{aligned} & S_{6} > S_{5}+1 \\\\ & \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\ & a r^{5} > 1 \\\\ & \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\ & \Rightarrow \quad m^{2} < 500 \\\\ & \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\ & \quad S_{7} < S_{6}+\frac{1}{2} \\\\ & \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\ & \quad a r^{6}<\frac{1}{2} \end{aligned} $$
$\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$
$$ \frac{1}{m^{3}}<\frac{1}{1000} $$
$\Rightarrow m \in(10, \infty)$
Possible values of $m$ is $\{11,12,....22 \}$
$\because m \in N$
Total 12 values
$$ a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}} $$
Now,
$$ \begin{aligned} & S_{6} > S_{5}+1 \\\\ & \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\ & a r^{5} > 1 \\\\ & \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\ & \Rightarrow \quad m^{2} < 500 \\\\ & \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\ & \quad S_{7} < S_{6}+\frac{1}{2} \\\\ & \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\ & \quad a r^{6}<\frac{1}{2} \end{aligned} $$
$\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$
$$ \frac{1}{m^{3}}<\frac{1}{1000} $$
$\Rightarrow m \in(10, \infty)$
Possible values of $m$ is $\{11,12,....22 \}$
$\because m \in N$
Total 12 values
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