1. Understand the problem:
We are given a geometric progression (G.P.) with first term 'a' and common ratio 'r'. We need to find the ratio of the sum of the first n terms (\( S_n \)) to the sum of the first 2n terms (\( S_{2n} \)).
2. Recall the formula for sum of a G.P.:
The sum of the first n terms of a G.P. is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \quad \text{(for \( r \neq 1 \))} \]
Similarly, the sum of the first 2n terms is:
\[ S_{2n} = a \frac{1 - r^{2n}}{1 - r} \]
3. Compute the ratio \( \frac{S_n}{S_{2n}} \):
Divide \( S_n \) by \( S_{2n} \):
\[ \frac{S_n}{S_{2n}} = \frac{a \frac{1 - r^n}{1 - r}}{a \frac{1 - r^{2n}}{1 - r}} = \frac{1 - r^n}{1 - r^{2n}} \]
4. Simplify the expression:
Notice that \( 1 - r^{2n} \) can be factored as \( (1 - r^n)(1 + r^n) \). Thus:
\[ \frac{1 - r^n}{1 - r^{2n}} = \frac{1 - r^n}{(1 - r^n)(1 + r^n)} = \frac{1}{1 + r^n} \]
5. Match the result to the options:
The simplified form \( \frac{1}{1 + r^n} \) corresponds to option (B) \( \frac{1}{r^n + 1} \).
Correct Answer: (B) \( \frac{1}{r^n + 1} \)
The sum of the first term $S_1 = a$.
For two terms, $S_2 = a(1 + r)$. Then: $\frac{S_1}{S_2} = \frac{a}{a(1 + r)} = \frac{1}{r + 1}$.
Hence, $\frac{1}{r + 1}$ is the correct answer.
The function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is an increasing function in:
If \( A = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \), then \( A^{50} \) is:
The range of the function \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is equal to?