Note:

  • S_{\infty} = \frac{a}{1-r}

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

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

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

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

iv) 10 - 9 +8.1 - 7.29 + \ldots \infty

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:

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

Here a = 1        r = \frac{  \frac{-1}{3}  }{1} = \frac{-1}{3}

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

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

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

\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})

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

= ( \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)

= \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 )

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

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

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

iv)    10 - 9 +8.1 - 7.29 + \ldots \infty

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

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

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

= ( \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 )

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

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

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

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

\\

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

Answer:

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

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

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

= 9^{1/2}

= 3 = RHS. Hence proved.

\\

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

Answer:

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

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

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

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

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

\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

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

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

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

\Rightarrow S = 4

Now substituting back we get

= 2^{ \frac{1}{2^2} [ 4 ]  }

= 2^1 = 2 = RHS. Hence proved.

\\

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

Answer:

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

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

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.

\\

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 4 , and the difference between the third and fifth term is equal to 32/81 .

Answer:

Given  a = 4

Let the common ratio be r

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

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

Also given a_3 - a_5 = \frac{32}{81}

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

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

Let r^2 = x

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

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

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

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

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

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

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

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

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

\\

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

Answer:

We have 0.\overline{125}

= 0.125125125\ldots

= 0.125 + 0.000125 + 0.000000125 + \ldots

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

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

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

= \frac{125}{999}

\\

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

Answer:

0.4\overline{23}

= 0.4232323\ldots

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

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

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

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

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

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

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

= \frac{419}{990}

\\

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

Answer:

i)       0.\overline{3}

= 0.33333\ldots

= 0.3 + 0.03 + 0.003 + \ldots

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

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

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

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

ii)      0.\overline{231}

= 0.231231231\ldots

= 0.231 + 0.000231 + 0.000000231 + \ldots

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

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

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

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

iii)     3.5\overline{2}

= 3.522222\ldots

= 3.5 + 0.02 + 0.002 + 0.0002 + \ldots

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

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

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

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

 iv)   0.6\overline{8}

= 0.688888\ldots

= 0.6 + 0.08 + 0.008 + 0.0008 + \ldots

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

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

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

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

\\

Question 9: One side of an equilateral triangle is 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 = 18

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

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

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

i) Sum of the perimeters of the triangles

= 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 = \frac{1}{2} \times base \times height .

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

\therefore the sum of the area of the triangles

= \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

\\

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

Answer:

Let the first term be a and the common ratio be r

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

Given a = \frac{ar}{1-r}

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

\Rightarrow 1-r = r

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

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

\\

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

Answer:

Let the first term be a and the common ratio be r

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

Given a + ar = 5

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

\Rightarrow 1-r = 3r

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

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

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

\\

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

Answer:

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

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

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

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

and \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.

\\

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

Answer:

Let the first term be a and the common ratio be r .

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

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

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}

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

S^2 - rS^2 = S_1 + r S_1

S^2 - S_1 = r ( S_1 + S^2)

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

Substituting in i)

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

\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)

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