Class 11: Geometric Progression – Exercise 20.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note:

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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 $\displaystyle S_p + s_p = 2 S_{2p}$

Answer:

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

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

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

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 $\displaystyle 32/81 .$

Answer:

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

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.

Answer:

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 $\displaystyle 0.4\overline{23}$

Answer:

$\displaystyle 0.4\overline{23}$

$\displaystyle = 0.4232323\ldots$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle = 0.33333\ldots$

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

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

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

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

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

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

$\displaystyle = 0.231231231\ldots$

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

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

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

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

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

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

$\displaystyle = 3.522222\ldots$

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

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

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

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

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

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

$\displaystyle = 0.688888\ldots$

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

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

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

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

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

$\displaystyle \\$

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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