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  • The 4th term of GP is 500 and its common ratio is <img alt="\small \frac{1}{m}, m \in \mathrm{N} ." src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20%5Csmall%20%5Cfrac%7B

The 4th term of GP is 500 and its common ratio is \small \frac{1}{m}, m \in \mathrm{N} . Let \small S_{\mathrm{n}} denote the sum of the first n

terms of this\small \text { GP. If } \mathrm{S}_6>\mathrm{S}_5+1 \text { and } \mathrm{S}_7<\mathrm{S}_6+\frac{1}{2}, then the number of possible values of m is

Option: 1

12


Option: 2

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Option: 3

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Option: 4

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Answers (1)

\begin{aligned} & \mathrm{T}_4=500 \Rightarrow \mathrm{a}\left(\frac{1}{\mathrm{~m}}\right)^3=500 \Rightarrow \mathrm{a}=500 \mathrm{~m}^3 \\ & \text { Now } \mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}=\mathrm{a}\left(\frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}\right)-\mathrm{a}\left(\frac{1-\mathrm{r}^{\mathrm{n}-1}}{1-\mathrm{r}}\right) \end{aligned}

\begin{aligned} & =\frac{a}{1-r}\left[r^{n-1}(1-r)\right] \\ & =a r^{n-1} \\ & =500 \mathrm{~m}^3\left(\frac{1}{m}\right)^{n-1} \\ & S\!_n-S_{n-1}=500 \mathrm{~m}^{4-n} \end{aligned}

\!\!\!\!\!\!\!\!Now\; S_6-S_s>1 \Rightarrow 500 \mathrm{~m}^{-2}>1 \ldots (1)\\ \&\; S_7-S_6<\frac{1}{2} \Rightarrow 500 \mathrm{~m}^{-3}<\frac{1}{2} \ldots(2)

\left.\begin{array}{lc} \text { from(1) } & \mathrm{m}^2<500 \\ \text { from(2) } & \mathrm{m}^3>1000 \end{array}\right] 10<\mathrm{m} \leq 22

Number of possible values of m is = 12

Posted by

Ramraj Saini

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