Question 25 and 26, Exercise 4.7
Solutions of Question 25 and 26 of Exercise 4.7 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 25
Sum to $n$ term of the series (arithmetico-geometric series): $1+\frac{4}{7}+\frac{7}{7^{2}}+\frac{10}{7^{3}}+\ldots$
Solution.
The given arithmetic-geometric series is: \[ 1 + \frac{4}{7} + \frac{7}{7^2} + \frac{10}{7^3} + \ldots \]
The numbers \(1, 4, 7, 10, \ldots\) are in AP with \(a = 1\) and \(d = 3\).
The numbers \(1, \frac{1}{7}, \frac{1}{7^2}, \frac{1}{7^3}, \ldots\) are in GP with first term \(1\) and \(r = \frac{1}{7}\).
The sum of the first \(n\) terms of the arithmetico-geometric series is given by:
\begin{align*}
S_n& = \frac{a}{1 - r} + d \cdot r \cdot \frac{1 - r^n}{(1 - r)^2} - \frac{(a + n \cdot d) r^n}{1 - r}\\
&= \frac{1}{1 - \frac{1}{7}} + 3 \cdot \frac{1}{7} \cdot \frac{1 - \left(\frac{1}{7}\right)^n}{\left(1 - \frac{1}{7}\right)^2} - \frac{(1 + n \cdot 3) \left(\frac{1}{7}\right)^n}{1 - \frac{1}{7}}\\
&= \frac{7}{6} + \frac{21(1 - \frac{1}{7^n})}{36} - \frac{7(1 + 3n)}{6 \cdot 7^n}\\
&=\frac{7}{6} + \frac{7(7^n - 1)}{12 \times7^n } - \frac{1 + 3n}{6 \times 7^n}\\
&=\frac{2\times 7^{n+1}+7^{n+1}-7-2-6n}{12\times 7^n} \\
&=\frac{3 \times7^{n+1}-9-6n}{12\times 7^n}\\
&=\frac{3(7^{n+1}-3-2n)}{12 \times7^n}\\
&=\frac{7^{n+1}-3-2n}{4\times 7^n}\\
\end{align*}
This is required sum. 
Question 26
Sum to $n$ term of the series (arithmetico-geometric series): $1+\frac{7}{2}+\frac{13}{4}+\frac{19}{8}+\frac{25}{16}+\ldots$
Solution.
The given arithmetic-geometric series is: \[ 1 + \frac{7}{2} + \frac{13}{4} + \frac{19}{8} + \frac{25}{16} + \ldots \]
The numbers \(1, 7, 13, 19, 25, \ldots\) are in AP with \(a = 1\) and \(d = 6\).
The numbers \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\) are in GP with first term \(1\) and \(r = \frac{1}{2}\)
The sum of the first \(n\) terms of the arithmetico-geometric series is given by: \begin{align*} S_n &= \frac{a}{1 - r} + d \cdot r \cdot \frac{1 - r^n}{(1 - r)^2} - \frac{(a + n \cdot d) r^n}{1 - r}\\ &= \frac{1}{1 - \frac{1}{2}} + 6 \cdot \frac{1}{2} \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\left(1 - \frac{1}{2}\right)^2} - \frac{(1 + n \cdot 6) \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}}\\ & = 2 + 12(1 - \frac{1}{2^n}) - \frac{2(1 + 6n)}{2^n}\\ & = 2 + 12- \frac{12}{2^n} - \frac{(2 + 12n)}{2^n}\\ & = 14 - \frac{14 + 12n}{2^n}\\ & = \frac{14\times 2^n-14 + 12n}{2^n}\\ & = 2^{-n}(14\times 2^n-14 + 12n)\\ & = 2^{1-n}(7\times 2^n-7 + 6n)\end{align*}
This is the required sum. 
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