JEE Mains · Maths · STD 11 - 8. sequence and series
The \(4^{\text {tht }}\) term of \(GP\) is \(500\) and its common ratio is \(\frac{1}{m}, m \in N\). 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 \(15\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
\(T_4=500 \quad\) where \(a=\) first term, \(r =\) common ratio \(=\frac{1}{ m }, m \in N\) \(a r^3=500\) \(\frac{a}{m^3}=500\) \(S_n-S_{n-1}=a r^{n-1}\) \(S _6 > S _5+1 \quad\) and \(S _7- S _6 < \frac{1}{2}\) \(S _6- S _5 > 1 \quad \frac{ a }{ m ^6} < \frac{1}{2}\)…
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