JEE Mains · Maths · STD 11 - 8. sequence and series
If for \(x, y \in {R}, x>0,\) \(y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots . .\) upto \(\infty\) terms and \(\frac{2+4+6+\ldots+2 \mathrm{y}}{3+6+9+\ldots+3 \mathrm{y}}=\frac{4}{\log _{10} \mathrm{x}}\), then the ordered pair \((x, y)\) is equal to :
- A \(\left(10^{6}, 6\right)\)
- B \(\left(10^{4}, 6\right)\)
- C \(\left(10^{2}, 3\right)\)
- D \(\left(10^{6}, 9\right)\)
Answer & Solution
Correct Answer
(D) \(\left(10^{6}, 9\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{2(1+2+3+\ldots+y)}{3(1+2+3+\ldots+y)}=\frac{4}{\log _{10} x}\) \(\Rightarrow \log _{10} x=6 \Rightarrow x=10^{6}\) Now, \(y=\left(\log _{10} x\right)+\left(\log _{10} x^{\frac{1}{3}}\right)+\left(\log _{10} x^{\frac{1}{9}}\right)+. . \infty\)…
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