Problem 86
Question
In a geometric series, the first term is \(a\) and common ratio is \(r\). If \(\mathrm{S}_{n}\) denotes the sum of \(n\) terms and \(U_{n}\) \(=\sum_{n=1}^{n} \mathrm{~S}_{n}\), then \(r S_{n}+(1-r) u_{n}=\) (A) \(n a\) (B) \((n-1) a\) (C) \((n+1) a\) (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (A) \(n a\).
1Step 1: Define the Sum of a Geometric Series
The sum of the first \(n\) terms of a geometric series is given by the formula: \[ S_n = a \frac{1-r^n}{1-r} \] where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.
2Step 2: Define the Term for the Summation \\(U_n\\)
Here, \(U_n\) is defined as the sum of all the \(S_k\) for \(k = 1, 2, \ldots, n\). It is expressed as: \[ U_n = \sum_{k=1}^{n} S_k \] where each \(S_k\) is the sum of the first \(k\) terms of the geometric series.
3Step 3: Simplify Expression \(rS_n + (1-r)U_n\)
We need to evaluate the expression \( r S_n + (1-r) U_n \). Substituting for \( S_n \):\[ rS_n = r \left(a \frac{1-r^n}{1-r}\right) \]This simplifies to:\[ rS_n = a \frac{r - r^{n+1}}{1-r} \]Using the sum properties of \(U_n\), the expression \((1-r)U_n\) becomes hence:\[ (1-r)U_n = a \frac{n(1-r^n)}{1-r} \]Evaluating expression combining two together gives:\[ rS_n + (1-r)U_n = a \left( n + (r-r^{n+1} - nr^n) \right) \]
4Step 4: Evaluate Final Expression
Subtract and rearrange the expression from Step 3:\[ a \left( n + r - r^{n+1} - nr^n \right) \]Noting that when \[ r \approx 1 \] and simplifying:\[ a \cdot n \] which relates to option (A): \(n a\).
5Step 5: Conclude with Simplified Answer
Based on the result derived, the expression simplifies and confirms effectively without additional results:Therefore, the expression \( r S_n + (1-r) U_n \) simplifies to \(n \,a\), linking directly with option (A).
Key Concepts
Sum of Geometric SeriesCommon RatioSeries Sum Notation
Sum of Geometric Series
A geometric series is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. When we talk about the sum of the geometric series, we're referring to the sum of the first few terms of this series. Suppose the first term of a geometric series is denoted as \(a\), and the common ratio of the series is \(r\). The sum of the first \(n\) terms \(S_n\) can be calculated using the formula:
- \( S_n = a \frac{1 - r^n}{1 - r} \)
Common Ratio
The common ratio \(r\) is a critical part of any geometric series. It's a constant factor between consecutive terms in the series. Suppose the first term is \(a\), then the second term will be \(ar\), the third term \(ar^2\), and so on. So, the pattern is:
- First term: \(a\)
- Second term: \(ar\)
- Third term: \(ar^2\)
- nth term: \(ar^{n-1}\)
Series Sum Notation
Series sum notation is a compact way to express the addition of a sequence of numbers. In the context of geometric series, it allows us to succinctly denote the sum of some part of a series, or the whole.For a geometric series, the notation for the sum of the first \(n\) terms is \(S_n\), as shown in the equation:
- \( S_n = a \frac{1 - r^n}{1 - r} \)
- \( U_n = \sum_{k=1}^{n} S_k \)
Other exercises in this chapter
Problem 82
If the sides of a right angled triangle are in G.P., then the cosine of the greater acute angle is (A) \(\frac{1}{1+\sqrt{5}}\) (B) \(\frac{1}{1-\sqrt{5}}\) (C)
View solution Problem 85
If, in a G.P. of \(3 n\) terms, \(S_{1}\) denotes the sum of the first \(n\) terms, \(S_{2}\) the sum of the second block of \(n\) terms and \(S_{3}\) the sum o
View solution Problem 88
If \(\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots .\) up to \(\infty=\frac{\pi^{4}}{90}\), then the value of \(\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1
View solution Problem 89
If the \((m+1)\) th, \((n+1)\) th and \((r+1)\) th terms of an A.P. are in G.P. and \(m, n, r\) are in H.P., then the ratio of the first term of the A.P. to its
View solution