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 $..........$
- A$11$
- B$10$
- C$12$
- D$15$
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View SolutionLet $a_{1}, a_{2}, a_{3}, \dots$ be a $G$.$P$. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$. Then $a_{3} + a_{5} + a_{7}$ is equal to:
DifficultJEE MAIN 2026
View SolutionIf the $2^{\text{nd}}$ and $5^{\text{th}}$ terms of a $G$.$P$. are $24$ and $3$ respectively,then the sum of the $1^{\text{st}}$ six terms is:
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