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Class 11 RD Sharma – Chapter 20 Geometric Progressions- Exercise 20.4

  • Last Updated : 04 May, 2021

Question 1.1 Find the sum of the following series to infinity : 1-\frac13+\frac1{3^2}-\frac1{3^3}+\frac1{3^4}+......∞

Solution:

Given series: 1-\frac13+\frac1{3^2}-\frac1{3^3}+\frac1{3^4}+......∞

It is in the form of Geometric progression where first term=1 and common ratio=-1/3

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}   where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula



       S=\frac1{1-{(-\frac13)}}

         =\frac1{1+\frac13}

        =3/4

Thus, 1-\frac13+\frac1{3^2}-\frac1{3^3}+\frac1{3^4}+......∞=3/4

Question 1.2 Find the sum of the following series to infinity: 8+4\sqrt2+4+..........∞

Solution:

Given series 8+4\sqrt2+4+..........∞

It is in the form of Geometric progression where first term=8 and common ratio=1/√2

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}



where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

 S=\frac8{1-{\frac1{\sqrt2}}}

   =\frac{8\sqrt2}{\sqrt2-1}

  = 8√2(√2+1)

 =16+8√2

Thus, 8+4\sqrt2+4+..........∞    =16+8√2

Question 1.3 Find the sum of the following series to infinity :  \frac25+\frac3{5^2}+\frac2{5^3}+\frac3{5^4}+.......∞

Solution:

Given series \frac25+\frac3{5^2}+\frac2{5^3}+\frac3{5^4}+.......∞

(\frac25+\frac2{5^3}+......∞)+(\frac3{5^2}+\frac3{5^4}+.....∞)



These two equations are in the form of Geometric progression

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

S=(\frac{\frac25}{1-\frac1{25}}+\frac{\frac35}{1-\frac1{25}})

 =\frac{\frac25}{\frac{24}{25}}+\frac{\frac35}{\frac{24}{25}}

 =\frac1{\frac{24}{25}}

 =25⁄24

Thus, \frac25+\frac3{5^2}+\frac2{5^3}+\frac3{5^4}+.......∞=\frac{25}{24}

Question 1.4 Find the sum of the following series to infinity: 10-9+8.1-7.29+……….∞

 Solution:



  Given series: 10-9+8.1-7.29+……….∞

⇒  10-9+\frac{81}{10}-\frac{729}{100}+........ ∞

The above series are in Geometric Progression with first term=10 and common ratio=-9⁄10

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

⇒ S=\frac{10}{1-(-\frac{9}{10})}

⇒ S=100⁄19

Thus 10-9+8.1-7.29+..........∞=\frac{100}{19}

Question 1.5 Find the sum of the following series to infinity: \frac13+\frac1{5^2}+\frac1{3^3}+\frac1{5^4}+\frac1{3^5}+\frac1{5^6}+....∞

Solution:



Given series: \frac13+\frac1{5^2}+\frac1{3^3}+\frac1{5^4}+\frac1{3^5}+\frac1{5^6}+....∞

         ⇒ (\frac13+\frac1{3^3}+\frac1{3^5}+....∞)+(\frac1{5^2}+\frac1{5^4}+....∞)

 These two equations are in the form of Geometric progression

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

          ⇒ (\frac{\frac13}{1-\frac1{3^2}}+\frac{\frac1{5^2}}{1-\frac1{5^2}})

         ⇒ (\frac{\frac13}{\frac8{3^2}}+\frac{\frac1{5^2}}{\frac{24}{5^2}})

        ⇒ 3⁄8+1⁄24

        ⇒ 10⁄24



Thus, \frac13+\frac1{5^2}+\frac1{3^3}+\frac1{5^4}+\frac1{3^5}+\frac1{5^6}+....∞=\frac{10}{24}

Question 2: Prove that (9^{\frac13}.9^{\frac19}.9^{\frac1{27}}......∞)

Solution:

Let us consider left hand side of the given equation

 ⇒ (9^{\frac13}.9^{\frac19}.9^{\frac1{27}}......∞)

 ⇒ 9^{\frac13+\frac19+\frac1{27}....∞}

   The equation \frac13+\frac19+\frac1{27}....∞    is in the form of Geometric progression where first term=1⁄3 and common ratio=1⁄3

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

   ⇒ 9^{\frac{\frac13}{1-\frac13}}



   ⇒ 9^{\frac12}

   ⇒ 3= Right hand side of the given equation

Thus we proved that (9^{\frac13}.9^{\frac19}.9^{\frac1{27}}......∞)=3

Question 3: Prove that (2^{\frac14}.4^{\frac18}.8^{\frac1{16}}.16^{\frac1{32}}...∞)=2

Solution:

Let us consider left hand side of the given equation

let S= 2^{\frac14}.4^{\frac18}.8^{\frac1{16}}.16^{\frac1{32}}...∞

S=2^{\frac14+\frac28+\frac3{16}+....∞}

Denoting the terms in power with x

x=\frac14+\frac28+\frac3{16}+.......∞                 eq-(1)

We can clearly see that this equation is neither in GP nor in AP. But the numerator of this equation is in AP and denominator in GP



let us multiply (1) with 1/2

\frac12x=\frac12[\frac14+\frac28+\frac3{16}+.....∞]                   eq-(2)

Let us subtract (2) from (1)

x-\frac{x}2=[\frac14+\frac28+\frac3{16}+....∞]-[\frac18+\frac2{16}+\frac3{32}+...∞]

    \frac{x}2=\frac14+(\frac28-\frac18)+(\frac3{16}-\frac2{16})+...∞

    \frac{x}2=\frac14+\frac18+\frac1{16}+...∞

      x=\frac12+\frac14+\frac18+....∞

This equation is in the form of Geometric progression where first term=1/2 and common ratio=1/2

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio



Thus, on substituting the value of a and r in that formula

⇒  x=\frac{\frac12}{1-\frac12}

⇒  x=\frac{\frac12}{\frac12}

⇒ x=1

⇒S=2=Right hand side of the equation

Thus we proved that (2^{\frac14}.4^{\frac18}.8^{\frac1{16}}.16^{\frac1{32}}...∞)=2

Question 4: If Sp denotes the sum of series 1+rp+r2p+…….∞ and sp the sum of the series 1 – rp+r2p+…….∞. Prove that Sp+sp=2S2p.

Solution:

Given Sp=1+rp+r2p+…….∞

The above equation is in Geometric progression where first term=1 and common ratio=rp

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio where r<1

Thus, on substituting the value of a and r in that formula

here |r|<1 ⇒|rp|<1

Sp=\frac1{1-r^p}                 eq-(1)

and  sp=1 – rp+r2p+…….∞

The above equation is in Geometric progression where first term=1 and common ratio=-rp

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio where r<1

Thus, on substituting the value of a and r in that formula

here |r|<1 ⇒|rp|<1

sp=\frac1{1+r^p}                  eq-(2)

on adding (1) and (2)

Sp+sp=\frac1{1-r^p}+\frac1{1+r^p}

         = \frac{1-r^p+1+r^p}{(1-r^p)(1+r^p)}

         =\frac2{1-(r^p)^2}

        =\frac2{1-r^{2p}}                        eq-(3)

From (1) We can write \frac1{1-r^{2p}}  =S2p

From (3), we can write that Sp+sp=2S2p

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

Solution:



Let a denote the first term of G.P and r be the common ratio

we know that nth term of a G.P is given by an=arn-1

Given a=4 and a5-a3=\frac{32}{81}

Here a5=4r4, a3=4r2

⇒ 4r4-4r2=32⁄81

⇒  4r2(r2-1)=32⁄81

⇒ r2(r2-1)=8⁄81

let x=r2

⇒ x(x-1)=8⁄81

⇒ 81x2-81x-8=0

by using the formula of quadratic equation to solve the above equation, we get 

 x= \frac{-(-81)\pm\sqrt{(81)^2-4(-8)(-81)}}{162}

⇒ x=\frac{81\pm\sqrt{6561-2592}}{162}

⇒ x=18⁄162 or x=144⁄162

⇒ x=1⁄9  or x=8⁄9

⇒ r2=1⁄9  or r2=8⁄9

⇒ r=1⁄3  or r=2√2/3

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Let us take 1st case a=4 and r=1⁄3

Then, S=\frac4{1-\frac13}

⇒ S=6

Now, in 2nd case a=4 and r=\frac{2\sqrt{2}}3

Then, S=\frac4{1-\frac{2\sqrt2}3}

⇒ S=\frac{12}{3-2\sqrt2}

Thus the sum of the terms of an infinite decreasing G.P in which all the terms are positive are 6 and \frac{12}{3-2\sqrt2}

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

Solution:

Let x=0.125125125……            eq-(1)

Here 125 is the repeating term 

so,let us multiply both sides of the equation with a number such that complete repetitive part of number comes after decimal.



let us multiply (1) with 1000 in both sides ,we get

1000x=125.125125125……               eq-(2)

let us subtract (1) from (2)

1000x-x=125.0000

999x=125

   x=125⁄999

Thus, 0.125125125…. can be expressed as 125/999

Question 7: Find the rational number whose decimal expansion is 0.42\bar{3}

Solution:

Let x=0.423333333…………             eq-(1)

here 3 is the repeating part .

So let us multiply both sides of equation (1) with a number such that complete repetitive part comes after decimal

Let us multiply (1) with 100

100x =42.33333……….          eq-(2)

let us subtract (1) from (2) 

100x-x=(42.3333333………..)-(0.4233333….)

  99x =41.91

     x=41.91/99

   x=4191/9900

The rational number whose decimal expansion is 0.42\bar{3}   is 4191/9900

Question 8-1: Find the rational number having the following decimal expansion is 0.\bar3

Solution:

let x=0.333333333333………..

    x=0.3+0.03+0.003+………∞

   x=3(0.1+0.01+0.001+……..∞)

  x=3(\frac1{10}+\frac1{100}+\frac1{1000}+......∞)

The above equation is in Geometric Progression with common ratio=1⁄10 and first term=1⁄10

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

  x=3(\frac{\frac1{10}}{1-{\frac1{10}}})

  x=3×1⁄9

  x=1⁄3

The rational number having decimal expansion is 0.\bar3   is 1⁄3

Question 8-2: Find the rational number having the following decimal expansion is 0.\overline{231}

Solution:

let x=0.231231231……….          eq-(1)

     x=0.231+0.000231+0.000000231+……..∞

    x=231(0.001+0.00001+0.0000001+……∞)

    x=231(\frac1{10^3}+\frac1{10^6}+\frac1{10^9}+....∞)

The above equation is in the form of Geometric Progression with first term=1⁄1000 common ratio=1⁄1000

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio



Thus, on substituting the value of a and r in that formula

   x=231(\frac{\frac1{1000}}{1-\frac1{1000}})

     =231(\frac1{999})

     x=231/999

The rational number having decimal expansion is 0.\overline{231}   is 231/999

Question 8-3: Find the rational number having the following decimal expansion is 3.5\bar{2}

Solution:

let x=3.5222222……..

     x=3.5+0.02+0.002+0.0002+…….∞

     x=3.5+2(0.01+0.001+0.0001+…………∞)

let s= 0.01+0.001+0.0001+…………∞

    s=\frac1{100}+\frac1{1000}+\frac1{10000}+........∞

The above equation is in the form of Geometric Progression with first term=1⁄100 and common ratio=1⁄10

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

    s=\frac{\frac1{100}}{1-\frac1{10}}

     s =1⁄90

x=3.5+2s

  =3.5+2×1⁄90

  =35⁄10+1⁄45

  =(315+2)/90

  =317/90

The rational number having decimal expansion is 3.5\bar{2}   is 317/90.

Question 8-4: Find the rational number having the following decimal expansion is 0.6\bar8

Solution:

let x=0.688888……….

    x=0.6+0.08+0.008+0.0008+0.00008+…….∞

   x=0.6+8(0.01+0.001+0.0001+……..∞)

let s=0.01+0.001+0.0001+………∞

    s=\frac1{100}+\frac1{1000}+\frac1{10000}+......∞

The above equation is in the form of Geometric Progression with first term= 1⁄100 and common ratio=1⁄10  



We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

Thus, on substituting the value of a and r in that formula

 s=\frac{\frac1{100}}{1-(\frac1{10})}

  =1⁄90

x=0.6+8s

 =0.6+8×1⁄90

 =62⁄90

The rational number having decimal expansion is 0.6\bar8    is 62/90

Question 9: One side of an equilateral triangle is 18cm. The midpoints of the sides are joined to form another triangle whose midpoints, in turn, 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 the triangles.

Solution:

According to Mid-Point Theorem, the sides of each triangle formed by joining the midpoints of an equilateral triangle are half of the sides of the equilateral triangle.

Thus, the sides of equilateral triangles formed are 18cm,9cm,4.5cm,2.25cm,…………….

i) Sum of perimeters of all the triangles

Let Sum of perimeters of all the triangles be P

P=3×18+3×9+3×4.5+3×2.25+………….∞

  =3(18+9+9/2+9/4+……….∞)

It is in the form of G.P with first term=18 and common ratio=1⁄2

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

where a=first term and r=common ratio

P=3(\frac{18}{1-\frac12}  )

  =3(18×2)

  =3(36)=108

Thus, Sum of perimeters of all the triangles=108cms

ii) Sum of Areas of all the triangles

We know that the area of equilateral triangle =√3/4(a2)

where a is the length of the side of equilateral triangle.

Let Sum of Areas of all the triangles be A

A= √3/4(18)2+√3/4(9)2+√3/4(4.5)2+√3/4(2.25)2+…………….∞

   =√3/4((18)2+(9)2+(4.5)2+(2.25)2+…………∞)

It is in the form of G.P with first term=(18)2 and common ratio=1⁄4

We know that sum of G.P terms up to infinity is S=\frac{a}{1-r}

A= \frac{\sqrt3}4(\frac{18^2}{1-\frac14})

  = (√3/3)×182

  = 324/√3

 = 108√3 cm2

Thus the Sum of Areas of all the triangles =108√3cm2

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

Solution:

Given first term=1

let common ratio be r

Also given each term is the sum of all the terms which follow it

⇒ an=[an+1+an+2+an+3+……………∞] ∀ n∈ N

⇒ arn-1=arn+arn+1+arn+2+……………..∞           [since there are in G.P]

⇒ rn-1=rn+rn+1+rn+2+…………..∞

⇒ rn-1=\frac{r^n}{1-r}

⇒ rn-1(1-r)=rn

⇒ rn-1-rn=rn

⇒ rn-1=2rn

⇒ 1=2r

⇒ r=1⁄2

Now, the infinite series will be a,ar,ar2,ar3………..∞



And thus the infinite G.P is 1,1⁄2,1⁄4,1⁄8,………………….∞

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

Solution:

Let the first term be a and common difference be r in a given infinite G.P

Given a1+a2=5

⇒ a+ar=5     -(i)

Also, Given, an=3[an+1+an+2+an+3+………∞] ∀ n∈ N

 ⇒ arn-1=3[arn+1-1+arn+2-1+arn+3-1+………..∞]

            =3[arn+arn+1+arn+2+………….∞]

           =3arn[1+r+r2+………….∞]

           =3arn\frac1{1-r}

  arn-1=3arn\frac1{1-r}

  ⇒ r-1=3⁄1-r

  ⇒ 1-r=3r

  ⇒ 1=4r

 ⇒ r=1⁄4

On substituting the value of r in (i)

 ⇒ a+a(1⁄4)=5

 ⇒ a(1+1⁄4)=5

 ⇒ a(5⁄4)=5

 ⇒ a=4

Now, the G.P terms will be a,ar,ar2,ar3,……………∞

Thus, the G.P terms will be 4,1,1⁄4,1⁄16…………∞

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

Solution:

Let us Consider a infinite G.P series whose terms are a1,a2,a3,a4,a5…………∞ and common ratio r(|r|<1).

Also, let us assume the sum of all terms following each term will be S1,S2,S3,S4………….∞

we know that 1+r+r2+r3+r4+……………..∞=1⁄1-r

Now,S1=\frac{a_2}{1-r}  =\frac{ar}{1-r}

S2=\frac{a_3}{1-r}  =\frac{ar^2}{1-r}

S3=\frac{a_4}{1-r}  =\frac{ar^3}{1-r}

⇒ a1/S1=\frac{a}{\frac{ar}{(1-r)}}    =(1-r)/r

⇒a2/S2=\frac{ar}{\frac{ar^2}{(1-r)}}    =(1-r)/r

⇒ a3/S3=\frac{ar^2}{\frac{ar^3}{(1-r)}}    =(1-r)/r

We can clearly observe that an/Sn=constant

Thus, we proved that ratio of each term to the sum of all the terms that follow it are constant.

Question 13: If S denotes the sum of infinite G.P. S1 denotes the sum of squares of its items, 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}

Solution:

Let us consider an infinite G.P series 

let a be the first term and r be the common ratio in G.P series.

Given

 a+ar+ar2+ar3+…………..∞=S

⇒ a(1+r+r2+r3+…………..∞)=S



⇒ a(1⁄1-r)=S                          eq-(1)

Also Given

S1=a2+a2r2+a2r4+…………..∞

   =a2(1+r2+r4+………….∞)

   =a2(\frac1{1-r^2}  )

    =\frac{a}{1+r}  ×\frac{a}{1-r}

   S1=S×(\frac{a}{1+r}  )                   eq-(2)

Divide (2) by (1)

S1/S= \frac{\frac{aS}{1+r}}{\frac{a}{1-r}}

       =\frac{S}{\frac{1+r}{1-r}}

  \frac{S_1}{S^2}=\frac{1-r}{1+r}     

⇒ (1-r)S2=(1+r)S1

⇒ S2-S1=r(S2+S1)

⇒ r=\frac{S^2-S_1}{S^2+S_1}

On substituting the value of r in (1)

S= \frac{a}{1-\frac{{S^2-S_1}}{{S^2+S_1}}}

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

 S= \frac{a(S^2+S_1)}{2S_1}

2SS1=a(S2+S1)

⇒ a=\frac{2SS_1}{S^2+S_1}

Thus, we proved that a=\frac{2SS_1}{S^2+S_1}   and r=\frac{S^2-S_1}{S^2+S_1}

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