KCET · Maths · Sequences and Series
\(n\)th term of the series \(1+\frac{3}{7}+\frac{5}{7^2}+\frac{1}{7^2}+\ldots\) is
- A \(\frac{2 n+1}{7^n}\)
- B \(\frac{2 n-1}{7^n}\)
- C \(\frac{2 n+1}{7^n-1}\)
- D \(\frac{2 n-1}{7^n-1}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 n-1}{7^n-1}\)
Step-by-step Solution
Detailed explanation
We have,
\(\begin{aligned} & 1+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots \\ & \frac{1}{70}+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots\end{aligned}\)
Numerator are in AP
So, \(T_n\) of \(\mathrm{AP}=1+(n-1)^2=2 n-1\)
Denominator are in GP
So, \(T_n\) of \(\mathrm{GP}=7^{n-1}\)
\(T_n\) of given series \(=\frac{2 n-1}{7^{n-1}}\)
\(\begin{aligned} & 1+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots \\ & \frac{1}{70}+\frac{3}{7}+\frac{5}{72}+\frac{7}{73}+\ldots\end{aligned}\)
Numerator are in AP
So, \(T_n\) of \(\mathrm{AP}=1+(n-1)^2=2 n-1\)
Denominator are in GP
So, \(T_n\) of \(\mathrm{GP}=7^{n-1}\)
\(T_n\) of given series \(=\frac{2 n-1}{7^{n-1}}\)
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