Note:

$\displaystyle S_{\infty} = \frac{a}{1-r}$

Question 1: Find the sum of the following series to infinity:

$\displaystyle \text{i) } 1 - \frac{1}{3} + \frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4} + \ldots \infty$

$\displaystyle \text{ii) } 8 + 4\sqrt{2} + 4 + \ldots \infty$

$\displaystyle \text{iii) } \frac{2}{5} +\frac{3}{5^2} +\frac{2}{5^3} +\frac{3}{5^4} + \ldots \infty$

$\displaystyle \text{iv) } 10 - 9 +8.1 - 7.29 + \ldots \infty$

$\displaystyle \text{v) } \frac{1}{3} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^4} +\frac{1}{3^5} + \frac{1}{5^6} + \ldots \infty$

$\displaystyle \text{i) } 1 - \frac{1}{3} + \frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4} + \ldots \infty$

$\displaystyle \text{Here } a = 1 r = \frac{ \frac{-1}{3} }{1} = \frac{-1}{3}$

$\displaystyle \therefore S = \frac{1}{1-(\frac{-1}{3})} = \frac{3}{4}$

$\displaystyle \text{ii) } 8 + 4\sqrt{2} + 4 + \ldots \infty$

$\displaystyle \text{Here } a = 8 r = \frac{ 4\sqrt{2} }{8} = \frac{1}{\sqrt{2}}$

$\displaystyle \therefore S = \frac{8}{1-(\frac{1}{\sqrt{2}})} = \frac{8\sqrt{2}}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = 8\sqrt{2} (\sqrt{2}+1) = 8(2 + 2\sqrt{2})$

$\displaystyle \text{iii) } \frac{2}{5} +\frac{3}{5^2} +\frac{2}{5^3} +\frac{3}{5^4} + \ldots \infty$

$\displaystyle = ( \frac{2}{5} + \frac{2}{5^3} +\frac{2}{5^5} + \ldots ) + ( \frac{3}{5^2} +\frac{3}{5^4} +\frac{3}{5^6} + \ldots)$

$\displaystyle = \frac{2}{5} ( 1 + \frac{1}{5^2} +\frac{1}{5^4} + \ldots ) + \frac{3}{5^2} ( 1+ \frac{1}{5^2} +\frac{1}{5^4} + \ldots )$

$\displaystyle = \frac{2}{5} ( \frac{1}{1 - \frac{1}{25}} ) + \frac{3}{25} ( \frac{1}{1 - \frac{1}{25}} )$

$\displaystyle = \frac{25}{24} ( \frac{2}{5} + \frac{3}{25})$

$\displaystyle = \frac{25}{24} \cdot \frac{13}{25} = \frac{13}{24}$

$\displaystyle \text{iv) } 10 - 9 +8.1 - 7.29 + \ldots \infty$

$\displaystyle \text{Here } a = 10 r = \frac{ -9 }{10} = 0.9$

$\displaystyle \therefore S = \frac{10}{1-0.9} = \frac{10}{1.9} = \frac{100}{19}$

$\displaystyle \text{v) } \frac{1}{3} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^4} +\frac{1}{3^5} + \frac{1}{5^6} + \ldots \infty$

$\displaystyle = ( \frac{1}{3}+ \frac{1}{3^3}+ \frac{1}{3^5} + \ldots \infty ) + ( \frac{1}{5^2}+ \frac{1}{5^4}+ \frac{1}{5^6} + \ldots \infty )$

$\displaystyle = \frac{\frac{1}{3}}{1 - \frac{1}{3^2}} + \frac{\frac{1}{5^2}}{1 - \frac{1}{5^2}}$

$\displaystyle = \frac{1}{3} ( \frac{9}{9-1} ) + \frac{1}{25} ( \frac{25}{25-1} )$

$\displaystyle = \frac{3}{8} + \frac{1}{24}$

$\displaystyle = \frac{10}{24} = \frac{5}{12}$

$\displaystyle \\$

Question 2: Prove that $\displaystyle ( 9^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 9^{\frac{1}{27}} \ldots \infty ) = 3$

$\displaystyle \text{LHS } = 9^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 9^{\frac{1}{27}} \ldots \infty$

$\displaystyle = 9^{ \big( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \infty \big)}$

$\displaystyle = 9^{\frac{1/3}{1 - 1/3} }$

$\displaystyle = 9^{1/2}$

$\displaystyle = 3 =$ RHS. Hence proved.

$\displaystyle \\$

Question 3: Prove that $\displaystyle ( 2^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 8^{\frac{1}{16}} \cdot {16}^{\frac{1}{32}} \ldots \infty ) = 2$

$\displaystyle 2^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 8^{\frac{1}{16}} \cdot {16}^{\frac{1}{32}} \ldots \infty$

$\displaystyle = 2^{ \big(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots \infty \big) }$

$\displaystyle = 2^{ \big( \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \ldots \infty \big) }$

$\displaystyle = 2^{ \big(\frac{1}{2^2} [ 1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots \infty ] \big) }$

$\displaystyle \text{Let } S = 1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots \infty$

$\displaystyle \therefore \frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \ldots \infty$

Subtracting the two we get

$\displaystyle \Rightarrow \frac{S}{2} = 1 + ( \frac{2-1}{2} ) + ( \frac{3-2}{2^2} ) + ( \frac{4-3}{2^3} ) + \ldots \infty$

$\displaystyle \Rightarrow \frac{S}{2} = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \ldots \infty$

$\displaystyle \Rightarrow \frac{S}{2} = \frac{1}{1-\frac{1}{2}}$

$\displaystyle \Rightarrow S = 4$

Now substituting back we get

$\displaystyle = 2^{ \frac{1}{2^2} [ 4 ] }$

$\displaystyle = 2^1 = 2 =$ RHS. Hence proved.

$\displaystyle \\$

Question 4: If $\displaystyle S_p$ denotes the sum of the series $\displaystyle 1+r^p+ r^{2p} + \ldots \infty$ and $\displaystyle s_p$ the sum of the series $\displaystyle 1-r^p+ r^{2p} - \ldots \infty$, prove that $\displaystyle S_p + s_p = 2 S_{2p}$

$\displaystyle \text{Given } S_p = 1+r^p+ r^{2p} + \ldots \infty = \frac{1}{1-r^p}$

$\displaystyle \text{Also } s_p = 1-r^p+ r^{2p} - \ldots \infty = \frac{1}{1 +r^p}$

$\displaystyle S_p + s_p = \frac{1}{1-r^p} + \frac{1}{1 +r^p} = \frac{1 + r^p + 1 - r^p}{1 - r^{2p}} = \frac{2}{1-r^{2p}} = 2 S_{2p}$

Hence proved.

$\displaystyle \\$

Question 5: Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is $\displaystyle 4$, and the difference between the third and fifth term is equal to $\displaystyle 32/81 .$

$\displaystyle \text{Given } a = 4$

Let the common ratio be $\displaystyle r$

$\displaystyle \therefore S = 4 + 4r + 4r^2 + \ldots \infty$

$\displaystyle \Rightarrow S = \frac{4}{1-r}$

$\displaystyle \text{Also } \text{Given } a_3 - a_5 = \frac{32}{81}$

$\displaystyle \Rightarrow ar^2 - ar^4 = \frac{32}{81}$

$\displaystyle \Rightarrow r^2 - r^4 = \frac{8}{81}$

$\displaystyle \text{Let } r^2 = x$

$\displaystyle \therefore x - x^2 = \frac{8}{81}$

$\displaystyle \Rightarrow 81 x - 81x^2 - 8 = 0$

$\displaystyle \Rightarrow 81x^2 - 81x + 8 = 0$

$\displaystyle \Rightarrow ( 9x - 8)(9x-1) = 0$

$\displaystyle \Rightarrow x = \frac{8}{9}, \frac{1}{9}$

$\displaystyle \text{When } x = \frac{8}{9} , \ \ \ \ \ \ r^2 = \frac{8}{9} \Rightarrow r = \frac{2\sqrt{2}}{3}$

$\displaystyle \text{When } x = \frac{1}{9} , \ \ \ \ \ \ r^2 = \frac{1}{9} \Rightarrow r = \frac{1}{3}$

$\displaystyle \text{When } r = \frac{2\sqrt{2}}{3} , \ \ \ \ \ \ S = \frac{4}{1 - \frac{2\sqrt{2}}{3} } = \frac{12}{3-2\sqrt{2}}$

$\displaystyle \text{When } r = \frac{1}{3} , \ \ \ \ \ \ \ \ \ \ S = \frac{4}{1-\frac{1}{3}} = 6$

$\displaystyle \\$

Question 6: Express the recurring decimal $\displaystyle 0.125125125...$ as a rational number.

We have $\displaystyle 0.\overline{125}$

$\displaystyle = 0.125125125\ldots$

$\displaystyle = 0.125 + 0.000125 + 0.000000125 + \ldots$

$\displaystyle = \frac{125}{10^3} + \frac{125}{10^6} + \frac{125}{10^9} + \ldots$

$\displaystyle = \frac{125}{10^3} \Big( 1 + \frac{1}{10^3} + \frac{1}{10^6} + \ldots \Big)$

$\displaystyle = \frac{125}{10^3} \Big( \frac{1}{1- \frac{1}{10^3}} \Big)$

$\displaystyle = \frac{125}{999}$

$\displaystyle \\$

Question 7: Find the rational number whose decimal expansion is $\displaystyle 0.4\overline{23}$

$\displaystyle 0.4\overline{23}$

$\displaystyle = 0.4232323\ldots$

$\displaystyle = \frac{1}{10} \Big[ 4.232323\ldots \Big]$

$\displaystyle = \frac{1}{10} \Big[ 4 + 0.23 + 0.0023 + 0.000023 + \ldots \Big]$

$\displaystyle = 0.4 + \frac{1}{10} \Big[ 0.23 + 0.0023 + 0.000023 + \ldots \Big]$

$\displaystyle = 0.4 + \frac{1}{10} \Big[ \frac{23}{10^2} + \frac{23}{10^4} + \frac{23}{10^6} + \ldots \Big]$

$\displaystyle = 0.4 + \frac{1}{10} \cdot \frac{23}{10^2} \Big[ 1 + \frac{1}{10^2} + \frac{1}{10^4} + \ldots \Big]$

$\displaystyle = 0.4 + \frac{23}{10^3} \Big( \frac{1}{1- \frac{1}{10^2}} \Big)$

$\displaystyle = 0.4 + \frac{23}{10^3} \cdot \frac{10^2}{99}$

$\displaystyle = \frac{419}{990}$

$\displaystyle \\$

Question 8: Find the rational numbers having the following decimal expansions: $\displaystyle \text{i) } 0.\overline{3}$ $\displaystyle \text{ii) } 0.\overline{231}$ $\displaystyle \text{iii) } 3.5\overline{2}$ $\displaystyle \text{iv) } 0.6\overline{8}$

$\displaystyle \text{i) } 0.\overline{3}$

$\displaystyle = 0.33333\ldots$

$\displaystyle = 0.3 + 0.03 + 0.003 + \ldots$

$\displaystyle = \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \ldots$

$\displaystyle = \frac{3}{10} [ 1 + \frac{1}{10} + \frac{1}{10^2} + \ldots ]$

$\displaystyle = \frac{3}{10} \cdot ( \frac{1}{1- \frac{1}{10}} )$

$\displaystyle = \frac{3}{10} \cdot \frac{10}{9} = \frac{1}{3}$

$\displaystyle \text{ii) } 0.\overline{231}$

$\displaystyle = 0.231231231\ldots$

$\displaystyle = 0.231 + 0.000231 + 0.000000231 + \ldots$

$\displaystyle = \frac{231}{10^3} + \frac{231}{10^6} + \frac{231}{10^9} + \ldots$

$\displaystyle = \frac{231}{10^3} [ 1 + \frac{1}{10^3} + \frac{1}{10^6} + \ldots ]$

$\displaystyle = \frac{231}{10^3} \cdot ( \frac{1}{1- \frac{1}{10^3}} )$

$\displaystyle = \frac{231}{10^3} \cdot \frac{10^3}{999} = \frac{231}{999}$

$\displaystyle \text{iii) } 3.5\overline{2}$

$\displaystyle = 3.522222\ldots$

$\displaystyle = 3.5 + 0.02 + 0.002 + 0.0002 + \ldots$

$\displaystyle = 3.5 + \frac{2}{100} + \frac{2}{10^3} + \frac{2}{10^4} + \ldots$

$\displaystyle = 3.5 + \frac{2}{100} [ 1 + \frac{1}{10} + \frac{1}{10^2} + \ldots ]$

$\displaystyle = 3.5 + \frac{2}{100} \cdot ( \frac{1}{1- \frac{1}{10}} )$

$\displaystyle = 3.5 + \frac{2}{100} \cdot \frac{10}{9} = \frac{35}{10} + \frac{2}{90} = \frac{317}{90}$

$\displaystyle \text{iv) } 0.6\overline{8}$

$\displaystyle = 0.688888\ldots$

$\displaystyle = 0.6 + 0.08 + 0.008 + 0.0008 + \ldots$

$\displaystyle = 0.6 + \frac{8}{100} + \frac{8}{10^3} + \frac{8}{10^4} + \ldots$

$\displaystyle = 0.6 + \frac{8}{100} [ 1 + \frac{1}{10} + \frac{1}{10^2} + \ldots ]$

$\displaystyle = 0.6 + \frac{8}{100} \cdot ( \frac{1}{1- \frac{1}{10}} )$

$\displaystyle = 0.6 + \frac{8}{100} \cdot \frac{10}{9} = \frac{6}{10} + \frac{8}{90} = \frac{62}{90} = \frac{31}{45}$

$\displaystyle \\$

Question 9: One side of an equilateral triangle is $\displaystyle 18$ cm. The mid-points of its sides are joined to form another triangle whose mid-points, in tum, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.

We know, the length of the first triangle $\displaystyle = 18$

Length of the side of the second triangle $\displaystyle = \frac{1}{2} \times 18 = 9$

Length of the side of the third triangle $\displaystyle = \frac{1}{2} \times 9 = \frac{9}{2}$

Hence the series is $\displaystyle 18, 9, \frac{9}{2} , \ldots$

i) Sum of the perimeters of the triangles

$\displaystyle = 3 \Big[ 18 + 9 + \frac{9}{2} + \ldots \Big] = 3 \Big( \frac{18}{1 - \frac{1}{2}} \Big) = 108$ cm

ii) Area of the triangle $\displaystyle = \frac{1}{2} \times base \times height$.

We know, if the base of an equilateral triangle is $\displaystyle a$, then the area $\displaystyle = \frac{\sqrt{3}}{4} a^2$

$\displaystyle \therefore$ the sum of the area of the triangles

$\displaystyle = \frac{\sqrt{3}}{4} \Big[ 18^2 + 9^2 + ( \frac{9}{2} )^2 + \ldots \Big] = \frac{\sqrt{3}}{4} \Big( \frac{18^2}{1 - \frac{1}{4}} \Big) = 108 \sqrt{3} \ cm^2$

$\displaystyle \\$

Question 10: Find an infinite G.P. whose first term is $\displaystyle 1$ and each term is the sum of all the terms which follow it.

Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$

Therefore G.P. is $\displaystyle a, ar, ar^2, ar^3, \ldots$

$\displaystyle \text{Given } a = \frac{ar}{1-r}$

$\displaystyle \Rightarrow 1 = \frac{r}{1-r}$

$\displaystyle \Rightarrow 1-r = r$

$\displaystyle \Rightarrow r = \frac{1}{2}$

Therefore G.P. is $\displaystyle 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8} , \ldots$

$\displaystyle \\$

Question 11: The sum of first two terms of an infinite G.P. is $\displaystyle 5$ and each term is three times the sum of the succeeding terms. Find the G.P.

Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$

Therefore G.P. is $\displaystyle a, ar, ar^2, ar^3, \ldots$

$\displaystyle \text{Given } a + ar = 5$

$\displaystyle \Rightarrow a = 3 \Big( \frac{ar}{1-r} \Big)$

$\displaystyle \Rightarrow 1-r = 3r$

$\displaystyle \Rightarrow r = \frac{1}{4}$

$\displaystyle \text{Therefore } a = \frac{5}{1+\frac{1}{4}} = \frac{20}{5} = 4$

Therefore G.P. is $\displaystyle 4, 1, \frac{1}{4}, \frac{1}{16} , \ldots$

$\displaystyle \\$

Question 12: Show that in an infinite G.P. with common ratio $\displaystyle r(|r| < 1)$, each term bears a constant ratio to the sum of all terms that follow it.

Let the G.P. be $\displaystyle a, ar, ar^2, ar^3, \ldots$

$\displaystyle \text{Therefore } S_1 = \frac{ar}{1-r} S_2 = \frac{ar^2}{1-r} S_3 = \frac{ar^3}{1-r}$

$\displaystyle \text{Therefore } \frac{a_1}{S_1} = \frac{a}{\frac{ar}{1-r}} = \frac{1-r}{r}$

$\displaystyle \text{Similarly, } \frac{a_2}{S_2} = \frac{ar}{\frac{ar^2}{1-r}} = \frac{1-r}{r}$

and $\displaystyle \frac{a_3}{S_3} = \frac{ar^2}{\frac{ar^3}{1-r}} = \frac{1-r}{r}$

Therefore we see that each of the ratio is constant.

$\displaystyle \\$

Question 13: If $\displaystyle S$ denotes the sum of an infinite G.P. and $\displaystyle S_1$ denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively $\displaystyle \frac{2SS_1}{S^2 + S_1}$ and $\displaystyle \frac{S^2 - S_1}{S^2 + S_1} .$

Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$.

$\displaystyle \text{Given } S = a + ar + ar^2 + \ldots = \frac{a}{1-r}$ … … … … … i)

$\displaystyle S_1 = a^2 + a^2r^2 + a^2r^4 + \ldots = \frac{a^2}{1-r^2}$ … … … … … ii)

$\displaystyle \text{Now } S^2 = \frac{a^2}{(1-r)^2} = \frac{a^2 (1-r^2)}{(1-r^2)(1-r)^2} = S_1 \frac{1-r^2}{(1-r)^2} = S_1 \frac{1+r}{1-r}$

$\displaystyle \Rightarrow (1-r)S^2 = ( 1+r)S_1$

$\displaystyle S^2 - rS^2 = S_1 + r S_1$

$\displaystyle S^2 - S_1 = r ( S_1 + S^2)$

$\displaystyle r = \frac{S^2 - S_1}{S_1 + S^2}$

Substituting in i)

$\displaystyle S= a \cdot \Bigg( \frac{1}{1 - \frac{S^2 - S_1}{S_1 + S^2}} \Bigg)$

$\displaystyle \Rightarrow S = a \Big( \frac{S_1 + S^2}{(S_1+S^2) - ( S^2 - S_1)} \Big) = a \Big( \frac{S_1 + S^2}{2S_1} \Big)$

$\displaystyle \Rightarrow a = \frac{2 S S_1}{S_1 + S^2}$