Assertion (A) : If x+x22+x33+……∞=log76 then x=17Reason(R) : If x<1, then x+x22+x33+……∞=log11-x

x+x22+x33+.......∞ =-log (1-x)=log 76 ⇒ log 11-x=log 76 ⇒ 11-x=76 ⇒ 1-x=67 ⇒ x=17Both (A), (R) one time and R is correct explanation of A

Assertion (A) : If x+x22+x33+……∞=log76 then x=17Reason(R) : If x<1, then x+x22+x33+……∞=log11-x

x+x22+x33+.......∞ =-log (1-x)=log 76 ⇒ log 11-x=log 76 ⇒ 11-x=76 ⇒ 1-x=67 ⇒ x=17Both (A), (R) one time and R is correct explanation of A

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Assertion (A) : If $x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots \dots \infty = \log (\frac{7}{6}) then x = \frac{1}{7}$
Reason(R) : If $|x| < 1, then x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots \dots \infty = \log (\frac{1}{1 - x})$

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A is true, R is true and R is correct explanation of A

A is true, R is true and R is not correct explanation of A

A is true, R is false

A is false, R is true

answer is A.

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Detailed Solution

$\begin{array}{rcl} x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + . . . . . . . \infty \\ = & - \log (1 - x) = \log \frac{7}{6} \\ \Rightarrow & \log \frac{1}{1 - x} = \log \frac{7}{6} \\ \Rightarrow & \frac{1}{1 - x} = \frac{7}{6} \\ \Rightarrow & 1 - x = \frac{6}{7} \\ \Rightarrow & x = \frac{1}{7} \end{array}$