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QJEEM23Mab3d4: JEE Main 2023 Mathematics - Class 11 JEEM23Mab3d4 Solution: The 4 term of GP is 500 and its common ratio is . Let denote…

The 4 th^\mathrm{th} term of GP is 500 and its common ratio is 1m,m∈N\frac{1}{m},m\in\mathbb{N} . Let Sn\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 ___________

Step-by-Step Verified Solution

T4=500T_{4}=500

ar3=500⇒a=500r3a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}}

Now,

amp;S6gt;S5+1amp;a(1−r6)1−r−a(1−r5)1−rgt;1amp;ar5gt;1amp; Now, r=1m and a=500r3amp;⇒m2lt;500amp;∵mgt;0⇒m∈(0,105)amp;S7lt;S6+12amp;a(1−r6)1−rlt;a(1−r6)1−r+12amp;ar6lt;12\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}

∵r=1m\because r=\frac{1}{m} and a=500r5a=\frac{500}{r^{5}}

\frac{1}{m^{3}}<\frac{1}{1000}

⇒m∈(10,∞)\Rightarrow m \in(10, \infty)

Possible values of mm is {11,12,....22}\{11,12,....22 \}

∵m∈N\because m \in N

Total 12 values

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