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JEE Mains · Maths · STD 11 - 8. sequence and series

यदि \(x , y \in R , x >0\), के लिए \(y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots .\) अनंत पदों तक तथा \(\frac{2+4+6+\ldots+2 y }{3+6+9+\ldots+3 y }=\frac{4}{\log _{10} x }\) हैं, तो क्रमित युग्म \(( x , y )\) बराबर है

  1. A \(\left(10^{6}, 6\right)\)
  2. B \(\left(10^{4}, 6\right)\)
  3. C \(\left(10^{2}, 3\right)\)
  4. D \(\left(10^{6}, 9\right)\)
Verified Solution

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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