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

  • Last Updated : 19 Apr, 2021

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

(i) 1 – 1/3 + 1/32 – 1/33 + 1/34 + … ∞

Solution:

Given series is an infinite G.P. with first term(a) = 1 and common ratio(r) = –1/3.

We know sum of a G.P. up to infinity is given by, S = a/(1 – r).

= 1/[1 – ( – 1/3)]

= 1/(4/3)



= 3/4

Therefore, sum of the series to infinity is 3/4.

(ii) 8 + 4√2 + 4 + …. ∞

Solution:

Given series is an infinite G.P. with first term(a) = 8 and common ratio(r) = 4√2/8 = 1/√2 

We know sum of a G.P. up to infinity is given by, S = a/(1 – r).

= 8/[1 – (1/√2)]

= 8/[(√2 – 1)/√2]

= 8√2/(√2 – 1)

\frac{8\sqrt{2}(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)}

\frac{8(2 + \sqrt{2})}{2-1}

= 8(2 + √2)

Therefore, sum of the series to infinity is 8(2 + √2).

(iii) 2/5 + 3/52 + 2/53 + 3/54 + …. ∞

Solution:

Given series has sum, S = 2/5 + 3/52 + 2/53 + 3/54 + …. ∞ 

=> S = (2/5 + 2/53 + … ∞) + (3/52 + 3/54 + … ∞)

We know sum of a G.P. up to infinity is given by, S = a/(1–r).

Let S1 = 2/5 + 2/53 + … ∞

This is an infinite G.P. with first term(a) = 2/5 and common ratio(r) = 1/52 = 1/25.



So, S1\frac{\frac{2}{5}}{1-\frac{1}{25}}

\frac{\frac{2}{5}}{\frac{24}{25}}

\frac{10}{24}

Let S2 = 3/52 + 3/54 + … ∞

This is an infinite G.P. with first term(a) = 3/5 and common ratio(r) = 1/52 = 1/25.

So, S2\frac{\frac{3}{5}}{1-\frac{1}{25}}

\frac{\frac{3}{5}}{\frac{24}{25}}

\frac{3}{24}

Now, required sum, S = S1 + S2

\frac{10}{24} + \frac{3}{24}



= 13/24

Therefore, sum of the series to infinity is 13/24.

(iv) 10 – 9 + 8.1 – 7.29 + …. ∞

Solution:

Given series is an infinite G.P. with first term(a) = 10 and common ratio(r) = – 9/10

We know sum of a G.P. up to infinity is given by, S = a/(1 – r).

= 10/[1 – (–9/10)]

= 10/[1 + 9/10]

= 100/19

= 5.263

Therefore, sum of the series to infinity is 5.263.

(v) 1/3 + 1/52 + 1/33 + 1/54 + + 1/35 + 1/56 …. ∞

Solution:

Given series has sum, S = 1/3 + 1/52 + 1/33 + 1/54 + + 1/35 + 1/56 …. ∞

=> S = (1/3 + 1/33 + 1/35 … ∞) + (1/52 + 1/54 + 1/56 … ∞)

We know sum of a G.P. up to infinity is given by, S = a/(1–r).

Let S1 = 1/3 + 1/33 + 1/35 … ∞

This is an infinite G.P. with first term(a) = 1/3 and common ratio(r) = 1/32 = 1/9.

So, S1\frac{\frac{1}{3}}{1-\frac{1}{9}}

= 3/8

Let S2 = 1/52 + 1/54 + 1/56 … ∞

This is an infinite G.P. with first term(a) = 1/52 and common ratio(r) = 1/52 = 1/25.



So, S2\frac{\frac{1}{25}}{1-\frac{1}{25}}

= 1/24

Now, required sum, S = S1 + S2

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

= 10/24 

= 5/12

Therefore, sum of the series to infinity is 5/12.

Question 2. Prove that:

(9^{\frac{1}{3}}.9^{\frac{1}{9}}.9^{\frac{1}{27}}….∞) = 3.

Solution:

We can write the L.H.S. as,

9^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...∞}



We have S = 1/3 + 1/9 + 1/27 + . . . .∞ 

which forms an infinite G.P. with first term(a) = 1/3 and common ratio(r) = 1/3. 

Also, sum of a G.P. up to infinity is given by, S = a/(1–r).

\frac{\frac{1}{3}}{1 - \frac{1}{3}}

\frac{\frac{1}{3}}{\frac{2}{3}}

\frac{1}{2}

So, L.H.S. becomes, 9S = 91/2 = 3 = R.H.S.

Hence proved.

Question 3. Prove that:

(2^\frac{1}{4}.4^\frac{1}{8}.8^\frac{1}{16}.16^\frac{1}{32}...∞) = 2.

Solution:

The L.H.S. can be written as,



L.H.S. = (2^\frac{1}{4}.2^\frac{2}{8}.2^\frac{3}{16}.2^\frac{4}{32}...∞)

(2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}...∞})

We have, S = 1/4 + 2/8 + 3/16 + 4/32 + …∞  . . . . . (1)

Dividing both sides by 2, we get,

S/2 = 1/8 + 2/16 + 3/32 + …∞   . . . . (2)

Subtracting (2) from (1) we get,

\frac{S}{2} = \frac{1}{4} + (\frac{2}{8}-\frac{1}{8}) + (\frac{3}{16}-\frac{2}{16}) + (\frac{4}{32}-\frac{3}{32}) + ...∞

\frac{S}{2} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}+ ...∞

Now this is an infinite G.P. with first term(a) = 1/4 and common ratio(r) = 1/2.

Also, sum of a G.P. up to infinity is given by a/(1–r).



\frac{S}{2} = \frac{\frac{1}{4}}{1-\frac{1}{2}}

=> S/2 = 1/2

=> S = 1

So, L.H.S. becomes, 2S = 2 = R.H.S.

Hence proved.

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

Solution:

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

So, Sp = 1 + rp + r2p + … to ∞

This is an infinite G.P. with first term(a) = 1 and common ratio(r) = rp.

So, Sp = 1/(1 – rp)

Also, sp = 1 – rp + r2p – … to ∞

This is G.P. with first term(a) = 1 and common ratio(r) = –rp.

So, sp = 1/[1 – (-rp)] = 1/(1 + rp)

Thus, L.H.S. = sp + Sp

\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}}

Now, R.H.S. = 2 S2p

2\left[\frac{1}{1-r^{2p}}\right]

= L.H.S.



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:

Given G.P. has first term(a) = 4 and,

=> ar4 – ar2 = 32/81

=> 4(r4 – r2) = 32/81

=> r2(r2 – 1) = 8/81

Let’s suppose r2 = x, so the equation becomes,

=> x(x – 1) = 8/81

=> 81x2 – 81x – 8 = 0

Solving for x, we get,

=> x = \frac{-(-81)±\sqrt{81^2-4(-4)(81)}}{162}

=> x = \frac{81±63}{162}

=> x = 1/9 or x = 8/9

So, r2 = 1/9 or r2 = 8/9

=> r = 1/3 or r = 2√2/3

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

When a = 4 and r = 1/3, 

S = 4/(1 – (1/3))

= 4/(2/3)

= 6

When a = 4 and r = 2√2/3,

S = 4/[1 – (2√2/3)]

= 12/(3 – 2√2)

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

Solution:

We are given, 0.125125125 = 0.125 + 0.000125 + 0.000000125 + …

= 125/103 + 125/106 + 125/109 + …

This is an infinite G.P. with first term(a) = 125/103 and common ratio(r) = 1/103 = 1/1000.

 We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

S = \frac{125}{10^3}\left[\frac{1}{1-\frac{1}{1000}}\right]

 \frac{125}{1000}×\frac{1000}{999}



\frac{125}{999}

Therefore, the recurring decimal 0.125125125 can be expressed in rational number as 125/999.

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

Solution:

We are given, 0.4\overline{23}   = 0.4 + 0.0232323 . . . . . .

= 0.4 + 0.023 + 0.00023 + 0.0000023 . . . .

= 0.4 + 23/103 + 23/105 + 23/107 + . . . .

We have, S = 23/103 + 23/105 + 23/107 + . . . .

This is an infinite G.P. with first term(a) = 23/103 and common ratio(r) = 1/102 = 1/100.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

S = \frac{23}{10^3}\left[\frac{1}{1-\frac{1}{100}}\right]

\frac{23}{1000}×\frac{100}{99}

\frac{23}{990}

Thus, the rational number becomes, 

\frac{4}{10}+\frac{23}{990}

\frac{419}{990}

Therefore, 419/990 is the rational number for this decimal expansion.

Question 8. Find the rational numbers having the following decimal expansions:

(i) 0.\overline{3}

Solution:

We have, 0.\overline{3}   = 0.3333…

= 0.3 + 0.33 + 0.333 + . . . .

= 3/10 + 3/102 + 3/103 + . . . .



This is an infinite G.P. with first term(a) = 3/10 and common ratio(r) = 1/10.

We know, sum of a G.P. up to infinity is given by, S = a/(1–r).

\frac{3}{10}\left[\frac{1}{1-\frac{1}{10}}\right]

\frac{3}{10}×\frac{10}{9}

\frac{1}{3}

Therefore, 1/3 is the rational number for this decimal expansion.

(ii) 0.\overline{231}

Solution:

We have, 0.\overline{231}   = 0.231231231

= 0.231 + 0.000231 + 0.000000231 + . . . .

= 231/103 + 231/106 + 231/109 + . . . .

This is an infinite G.P. with first term(a) = 231/103 and common ratio(r) = 1/103 = 1/1000.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

\frac{231}{1000}\left[\frac{1}{1-\frac{1}{1000}}\right]

\frac{231}{1000}×\frac{1000}{999}

\frac{231}{999}

Therefore, 231/999 is the rational number for this decimal expansion.

(iii) 3.5\overline{2}

Solution:

We have, 3.5\overline{2}   = 3.522222

= 3.5 + 0.02 + 0.002 + 0.0002 + . . . .

= 3.5 + (2/102 + 2/103 + 2/104 + . . . ) 



We have, S = 2/102 + 2/103 + 2/104 + . . . 

This is an infinite G.P. with first term(a) = 2/102 and common ratio(r) = 1/10.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

So, S = \frac{2}{100}\left[\frac{1}{1-\frac{1}{10}}\right]

\frac{2}{100}×\frac{10}{9}

\frac{2}{90}

Thus, the number becomes,

\frac{35}{10}+\frac{2}{90}

\frac{317}{90}

Therefore, 317/90 is the rational number for this decimal expansion.

(iv) 0.6\overline{8}

Solution:

We have, 0.6\overline{8}   = 0.68888

= 0.6 + 0.08 + 0.888 + 0.8888 + . . . .

= 0.6 + (8/102 + 8/103 + 8/104 + . . . .)

We have, S = 8/102 + 8/103 + 8/104 + . . . .

This is an infinite G.P. with first term(a) = 8/102 and common ratio(r) = 1/10.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

S = \frac{8}{100}\left[\frac{1}{1-\frac{1}{10}}\right]

\frac{8}{100}×\frac{10}{9}

\frac{4}{45}



Thus, the number becomes,

\frac{3}{5}+\frac{4}{45}

\frac{31}{45}

Therefore, 31/45 is the rational number for this decimal expansion.

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 turn, are joined to form still another triangle. The process is continued infinitely. Find the sum of the:

(i) perimeters of all the triangles.

Solution:

The sides of all these triangles form an infinite G.P., 18, 9, 9/2, . . . .

Its first term is first term(a) = 18 and common ratio(r) = 9/18 = 1/2

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

Therefore, S = 18/[1 – 1/2]

= 18/(1/2)

= 36

Now, sum of perimeters of all the triangles, Sp = 3a1 + 3a2 + 3a3 + . . . ., 

where a1,a2,a3, . . . are sides of these triangles.

= 3 (a1 + a2 + a3 + . . . .)

= 3S

= 3(36)

= 108

Therefore, the sum of perimeters of all the triangles is 108 cm.

(ii) areas of all the triangles.

Solution:

Sum of areas of all the triangles, S_a = \frac{\sqrt{3}}{4}(18)^2+\frac{\sqrt{3}}{4}(9)^2+\frac{\sqrt{3}}{4}(\frac{9}{2})^2+...

\frac{\sqrt{3}}{4}\left[324 + 81 + \frac{81}{4} + ...\right]

Here the series 324 + 81 + (81/4)+ . . . 

forms an infinite G.P. with first term(a) = 324 and common ratio(r) = 81/324 = 1/4.

We know, sum of a G.P. up to infinity is given by, S = a/(1–r).

So, Sa\frac{\sqrt{3}}{4}\left[\frac{1}{1-\frac{1}{4}}\right]

\frac{\sqrt{3}}{4}×324×\frac{4}{3}

= 108√3

Therefore, the sum of areas of all the triangles is 108√3 cm2.

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 of G.P.(a) = 1 and,

an = an+1 + an+2 + an+3 + . . . .

We know nth term of a G.P. is given by an = arn-1. So, we get,

=> arn–1 = arn + arn+1 + arn+2 + . . . .

=> arn-1 = arn(1 + r + r2 + . . . .)

=> (1 + r + r2 + . . . .) = 1/r

Here the series on L.H.S forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

=> 1/(1 – r) = 1/r

=> 2r = 1

=> r = 1/2



As first term(a) is 1 and common ratio is 1/2, the required infinite G.P. is, 

1, 1/2, 1/4, 1/8, . . . 

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.

Solution:

Given a + ar = 5 => a(1 + r) = 5 . . . . (1)

Also, an = 3(an+1 + an+2 + an+3 + . . . .)

We know nth term of a G.P. is given by an = arn-1. So, we get,

=> arn-1 = 3(arn + arn+1 + arn+2 + . . . .)

=> arn-1 = 3arn(1 + r + r2 + . . . .)

=> (1 + r + r2 + . . . .) = 1/3r

Here the series on L.H.S forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

=> 1/(1 – r) = 1/3r

=> 4r = 1

=> r = 1/4

From (1), we get, a(1 + 1/4) = 5

=> a(5/4) = 5

=> a = 4

 As first term(a) is 4 and common ratio is 1/4, the required infinite G.P. is,

 4, 1, 1/4, 1/16, . . . .

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.

Solution:

According to the question, we have,

\frac{a_n}{(a_{n+1}+a_{n+1}+...∞)}   = \frac{ar^{n-1}}{ar^n+ar^{n+1}+...∞}

\frac{ar^{n-1}}{ar^n(1+r+r^2+...∞)}

Here the series in the denominator forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.

We know, sum of a G.P. up to infinity is given by, S = a/(1–r). So our expression becomes, 

\frac{ar^{n-1}}{ar^n(\frac{1}{1-r})}

\frac{1-r}{r}   

Since common ratio(r) of the G.P. remains constant, so value of the ratio \frac{1-r}{r}   also remains constant.

Therefore, in an infinite G.P., each term bears a constant ratio to the sum of all terms that follow it.

Question 13. If S denotes the sum of an infinite G.P., S1 denotes the sum of 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}

Solution:

We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).

Here, S = a + ar + ar2 + ar3 + . . . . = a/(1 – r) . . . . (1)

Also given that S1 = a2 + a2r2 + a2r4 + a2r6 + . . . .  = a2/(1 – r2)  . . . . (2)

Squaring both sides of (1), we get,

=> S^2 = \frac{a^2}{(1-r)^2}

From eq(2), we get,

=> S^2 = \frac{S_1(1-r^2)}{(1-r)^2}

=> S^2 = \frac{S_1(1+r)}{1-r}

=> S2 − S2r = S1 + S1r

=> (S1 + S2)r = S2 − S1



=> r = \frac{S^2-S_1}{S^2+S_1}

On putting value of r in eq(1), we get

=> a = S(1 − r)

=> a = S\left[1-\frac{S^2-S_1}{S^2+S_1}\right]

=> a = S\left[\frac{S^2+S_1-S^2+S_1}{S^2+S_1}\right]

=> a = \frac{2SS_1}{S^2+S_1}

Hence proved.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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