KCET · Maths · Sequences and Series
If \(S_n\) stands for sum to \(n\)-terms of a GP with \(a\) as the first term and \(r\) as the common ratio, then \(S_n: S_{2 n}\) is
- A \(r^n+1\)
- B \(\frac{1}{r^n+1}\)
- C \(r^n-1\)
- D \(\frac{1}{r^n-1}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{r^n+1}\)
Step-by-step Solution
Detailed explanation
\(S_n=\frac{a\left(1-r^n\right)}{1-r}\)
\(S_{2 n}=\frac{a\left(1-r^{2 n}\right)}{1-r}\)
\(\frac{S_n}{S_{2 n}}=\frac{a\left(1-r^n\right)}{1-r} \times \frac{1-r}{a\left(1-r^{2 n}\right)}=\frac{1-r^n}{\left(1+r^n\right)\left(1-r^n\right)}\)
Hence, \(S_n: S_{2 n}=\frac{1}{1+r^n}\)
\(S_{2 n}=\frac{a\left(1-r^{2 n}\right)}{1-r}\)
\(\frac{S_n}{S_{2 n}}=\frac{a\left(1-r^n\right)}{1-r} \times \frac{1-r}{a\left(1-r^{2 n}\right)}=\frac{1-r^n}{\left(1+r^n\right)\left(1-r^n\right)}\)
Hence, \(S_n: S_{2 n}=\frac{1}{1+r^n}\)
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