The 4^{text{th}} term of a GP is 500 and its | vedclass

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(C) Given $T_4 = ar^3 = 500$,where $r = \frac{1}{m}$.So,$a(\frac{1}{m})^3 = 500 \implies a = 500m^3$.Given $S_6 > S_5 + 1 \implies S_6 - S_5 > 1 \impl

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$12$

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(C) Given $T_4 = ar^3 = 500$,where $r = \frac{1}{m}$.So,$a(\frac{1}{m})^3 = 500 \implies a = 500m^3$.Given $S_6 > S_5 + 1 \implies S_6 - S_5 > 1 \implies T_6 > 1$.$ar^5 > 1 \implies 500m^3 \cdot (\frac{1}{m})^5 > 1 \implies \frac{500}{m^2} > 1 \implies m^2 Given $S_7 $ar^6  1000 \implies m > 10$.Combining $m^2  10$,we get $10 Since $m \in N$,$m \in \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22\}$.The number of possible values of $m$ is $12$.

The $4^{\text{th}}$ term of a $GP$ is $500$ and its common ratio is $\frac{1}{m}$,where $m \in N$. 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 $..........$

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