JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first \(20\) terms of the series \(\log _{\left(7^{\frac{1}{2}}\right)} x+\log _{\left(7^{\frac{1}{3}}\right)} x+\log _{\left(7^{\frac{1}{4}}\right)} x+\ldots\) is \(460,\) then \(x\) is equal to
- A \(7^{\frac{46}{21}}\)
- B \(7^{\frac{1}{2}}\)
- C \(e ^{2}\)
- D \(49\)
Answer & Solution
Correct Answer
(D) \(49\)
Step-by-step Solution
Detailed explanation
\(460=\log _{7} x \cdot(2+3+4+\ldots .+20+21)\) \(\Rightarrow 460=\log _{7} x \cdot\left(\frac{21 \times 22}{2}-1\right)\) \(\Rightarrow 460=230 \cdot \log _{7} x\) \(\Rightarrow \log _{7} x=2 \Rightarrow x=49\)
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