Question 29 and 30, Exercise 4.7

Solutions of Question 29 and 30 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.

Find sum to infinity of the series: $$1+4 x+7 x^{2}+10 x^{3}+\ldots$$

Solution.

The given arithmetic-geometric series is:
\[ 1 + 4x + 7x^2 + 10x^3 + \ldots \]

It can be rewritten as:
\[ 1 \times 1 + 4 \times x + 7 \times x^2 + 10 \times x^3 + \ldots \]

The numbers \(1, 4, 7, 10, \ldots\) are in AP with \(a = 1\) and \(d = 4 - 1 = 3\).

The numbers \(1, x, x^2, x^3, \ldots\) are in GP with first term \(1\) and \(r = x\).

The sum of the infinite arithmetico-geometric series is given by:
\[ S_{\infty} = \frac{a}{1 - r} + \frac{d r}{(1 - r)^{2}} \]

Thus, we have:
\begin{align*} S_{\infty} &= \frac{1}{1 - x} + \frac{(3 \times x)}{(1 - x)^{2}} \\ &= \frac{1}{1 - x} + \frac{3x}{(1 - x)^2}\\ &= \frac{1-x+3x}{(1 - x)^2}\\ &= \frac{1+2x}{(1 - x)^2} \end{align*}

This is the required sum.

Find sum to infinity of the series: $$3+\frac{6}{10}+\frac{9}{100}+\frac{12}{1000}+\ldots$$

Solution.

The given arithmetic-geometric series is:
\[ 3 + \frac{6}{10} + \frac{9}{100} + \frac{12}{1000} + \ldots \]

It can be rewritten as:
\[ 3 \times 1 + 6 \times \frac{1}{10} + 9 \times \frac{1}{100} + 12 \times \frac{1}{1000} + \ldots \]

\(3, 6, 9, 12, \ldots\) are in AP with \(a = 3\) and \(d = 6 - 3 = 3\).

\(1, \frac{1}{10}, \frac{1}{100}, \frac{1}{1000}, \ldots\) are in GP with first term \(1\) and \(r = \frac{1}{10}\).

The sum of the infinite arithmetico-geometric series is given by:
\[ S_{\infty} = \frac{a}{1 - r} + \frac{d r}{(1 - r)^{2}} \]

Thus, we have:
\begin{align*} S_{\infty} &= \frac{3}{1 - \frac{1}{10}} + \frac{(3 \times \frac{1}{10})}{\left(1 - \frac{1}{10}\right)^{2}} \\ &= \frac{3}{9/10} + \frac{3/10}{\left(9/10\right)^{2}} \\ &= \frac{10}{3} + \frac{10}{27} \\ &= \frac{100}{27} \end{align*}

This is the required sum.

< Question 27 & 28