Geometric progression

EXERCISE 20.3

QUESTION 1

Find the sum of the following geometric progressions:

(i) 2,6,18, \ldots to 7 terms;

Sol :

Here, a=2 and r=3

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

\therefore S_{7}=a\left(\frac{r^{7}-1}{r-1}\right)

=2\left(\frac{3^{7}-1}{3-1}\right)

= 2787 – 1

= 2186

 

(ii) 1,3,9,27, \ldots to 8 terms

Sol :

Here, a=1 and r=3

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

\therefore~S_{8}=a\left(\frac{r^{8}-1}{r-1}\right)

=1\left(\frac{3^{8}-1}{3-1}\right)

=\frac{6561-1}{2}

= 3280

 

(iii) 1,-1 / 2,1 / 4,-1 / 8, \ldots to 9 terms;

Sol :

Here, a=1 and r=-\dfrac{1}{2}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

\therefore S_{9}=a\left(\frac{1-r^{9}}{1-r}\right)

=1\left(\frac{1-\left(-\frac{1}{2}\right)^{9}}{1-\left(-\frac{1}{2}\right)}\right)

=\frac{1-\left(-\frac{1}{512}\right)}{\frac{3}{2}}

=\frac{\frac{513}{512}}{\frac{3}{2}}

=\frac{513 \times 2}{512 \times 3}

=\frac{171}{256}

 

(iv) \left(a^{2}-b^{2}\right),(a-b),\left(\frac{a-b}{a+b}\right), \ldots to n terms

Sol :

Here, a=a^{2}-b^{2} and r=\frac{1}{a+b}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

=\left(a^{2}-b^{2}\right)\left(\frac{1-\left(\frac{1}{a+b}\right)^{n}}{1-\left(\frac{1}{a+b}\right)}\right)

=\left(a^{2}-b^{2}\right)\left(\frac{\left(\frac{(a+b)^{n}-1}{(a+b)^{n}}\right)}{\frac{(a+b)-1}{a+b}}\right)

\Rightarrow S_{n}=\frac{(a+b)(a-b)}{(a+b)^{n-1}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right)

=\frac{(a-b)}{(a+b)^{n-2}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right)

 

(v) 4,2,1,1 / 2 \ldots to 10 terms.

Sol :

Here, a=4 and r=\frac{1}{2}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

=4\left(\frac{1-\left(\frac{1}{2}\right)^{10}}{1-\left(\frac{1}{2}\right)}\right)

=4\left(\frac{1-\left(\frac{1}{1044}\right)}{\frac{1}{2}}\right)

=8\left(1-\frac{1}{1024}\right)

=\frac{1023}{128}

 

QUESTION 2

Find the sum of the following geometric series:

(i) 0.15+0.015+0.0015+\ldots to 8 terms;

Sol :

Here, a=0.15 and r=\frac{a_{2}}{a_{1}}=\frac{0.015}{0.15} =\frac{1}{10}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

\mathrm{S}_{8}=a\left(\frac{1-r^{8}}{1-r}\right)

=0.15\left(\frac{1-\left(\frac{1}{10}\right)^{8}}{1-\frac{1}{10}}\right)

=0.15\left(\frac{1-\frac{1}{10^{8}}}{\frac{1}{10}}\right)

=\frac{1}{6}\left(1-\frac{1}{10^{8}}\right)

 

(ii) \sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2 \sqrt{2}}+\ldots to 8 terms

Sol :

Here, a=\sqrt{2} and r=\frac{1}{2}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

=\sqrt{2}\left(\frac{1-\left(\frac{1}{2}\right)^{8}}{1-\frac{1}{2}}\right)

=\sqrt{2}\left(\frac{1-\frac{1}{256}}{\frac{1}{2}}\right)

=2 \sqrt{2}\left(\frac{255}{256}\right)

=\frac{255 \sqrt{2}}{128}

 

(iii) \frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+\ldots to 5 terms;

Sol :

Here, a=\frac{2}{9} and r=-\frac{3}{2}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

S_{5}=a\left(\frac{r^{5}-1}{r-1}\right)

=\frac{2}{9}\left(\frac{\left(\frac{-3}{2}\right)^{5}-1}{\frac{-3}{2}-1}\right)

=\frac{2}{9}\left(\frac{\left(-\frac{243}{32}\right)-1}{\frac{-3}{\phantom{-}2}-1}\right)

=\frac{2}{9}\left(\frac{\frac{-275}{32}}{\frac{-5}{\phantom{-}2}}\right)

=\frac{1100}{1440}

=\frac{55}{72}

 

(iv) (x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots to n terms 

Sol :

 

(v) \frac{3}{5}+\frac{4}{5^{2}}+\frac{3}{5^{3}}+\frac{4}{5^{4}}+\ldots to 2n terms

Sol :

 

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

 

 

(vi) \frac{a}{1+i}+\frac{a}{(1+i)^{2}}+\frac{a}{(1+i)^{3}}+\ldots+\frac{a}{(1+i)^{n}}

Sol :

 

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

 

(vii) 1,-a_{t} a^{2},-a^{3}, \ldots to n terms (a \neq 1)

Sol :

 

 

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

 

(viii) x^{3}, x^{5}, x^{7}, \ldots to n terms

Sol :

 

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

 

 

(ix) \sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots to n terms

Sol :

 

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

 

QUESTION 3

Evaluate the following :

(i) \sum_{n=1}^{11}\left(2+3^{n}\right)

Sol :

S_{11}=\sum_{n=1}^{11}\left(2+3^{n}\right)

\Rightarrow S_{11}=\sum_{n=1}^{11} 2+\sum_{n=1}^{11} 3^{n}

\Rightarrow S_{11}=2 \times 11+\left(3+3^{2}+3^{3}+\ldots+3^{11}\right)

=22+3\left(\frac{3^{11}-1}{3-1}\right)

=22+\left(\frac{17747-1}{2}\right)

= 22 + 265719

= 265741

 

(ii) \sum_{k=1}^{n}\left(2^{k}+3^{k-1}\right)

Sol :

S_{n}=\sum_{k=1}^{n}\left(2^{k}+3^{k-1}\right)

=\sum_{k=1}^{n} 2^{k}+\sum_{k=1}^{n} 3^{k-1}

=\left(2+4+8+\ldots+2^{n}\right)+\left(1+3+9+\ldots+3^{n}\right)

=2\left(\frac{2^{n}-1}{2-1}\right)+1\left(\frac{3^{n}-1}{3-1}\right)

=\frac{1}{2}\left(2^{n+2}-4+3^{n}-1\right)

=\frac{1}{2}\left(2^{n+2}+3^{n}-5\right)

 

(iii) \sum_{n=2}^{10} 4^{n}

Sol :

\sum_{n=2}^{10} 4^{n}=4^{2}+4^{3}+4^{4}+\ldots+4^{10}

=16\left(\frac{4^{9}-1}{4-1}\right) =\frac{16}{3}\left(4^{9}-1\right)

 

QUESTION 4

Find the sum of the following series:

(i) 5+55+555+\ldots to n terms;

Sol :

taking 5 as common :

S_{n}=5[1+11+111+\ldots n terms ]

=\frac{5}{9}(9+99+999+\ldots . n terms )

=\frac{5}{9}\left\{(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots+\left(10^{n}-1\right)\right\}

=\frac{5}{9}\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1+\ldots n times )

=\frac{5}{9}\left\{10 \times \frac{\left(10^{n}-1\right)}{10-1}-n\right\}

=\frac{5}{9}\left\{\frac{10}{9}\left(10^{n}-1\right)-n\right\}

=\frac{5}{81}\left\{10^{n+1}-9 n-10\right\}

 

(ii) 7+77+777+\ldots to n terms

Sol :

Taking 7 as common :

S_{n}=7[1+11+111+\ldots n terms ]

=\frac{7}{9}(9+99+999+\ldots n terms )

=\frac{7}{9}\left\{(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots+\left(10^{n}-1\right)\right\}

=\frac{7}{9}\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1 \ldots n times )

=\frac{7}{9}\left\{10 \times \frac{\left(10^{n}-1\right)}{10-1}-n\right\}

=\frac{7}{9}\left\{\frac{10}{9}\left(10^{n}-1\right)-n\right\}

=\frac{7}{81}\left\{10^{n+1}-9 n-10\right\}

 

(iii) 9+99+999+\ldots to n terms

Sol :

This can be rewritten as :

=\left\{(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots+\left(10^{n}-1\right)\right\}

=\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1 \ldots n times )

=\left\{10 \times \frac{\left(10^{n}-1\right)}{10-1}-n\right\}

=\left\{\frac{10}{9}\left(10^{n}-1\right)-n\right\}

=\frac{1}{9}\left\{10^{n+1}-9 n-10\right\}

 

(iv) 0.5+0.55+0.555+\ldots to n terms.

Sol :

Taking 5 as common :

S_{n}=5[0.1+0.11+0.111+\ldots . n terms ]

=\frac{5}{9}(0.9+0.99+0.999+\ldots+ to n terms )

=\frac{5}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots\text{ n terms }\right\}

=\dfrac{5}{9}\left\{\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{100}\right)+\left(1-\dfrac{1}{1000}\right)+\dots\text{ n terms }\right\}

=\dfrac{5}{9}\left\{n-\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\dfrac{1}{10^3}+\dots{\text{ n terms }}\right)\right}

=\frac{5}{9}\left\{n-\frac{1}{10} \frac{\left(1-\left(\frac{1}{10}\right)^{n}\right)}{\left(1-\frac{1}{10}\right)}\right\}

=\frac{5}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right\}

 

(v) 0.6+0.66+0.666+\ldots . to n terms

Sol :

Taking 6 as common :

S_{n}=6[0.1+0.11+0.111+\ldots \text { n terms }]

=\frac{6}{9}(0.9+0.99+0.999+\ldots \text{n terms })

=\frac{6}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots\text{ n terms }\right\}

=\dfrac{6}{9}\left\{\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{100}\right)+\left(1-\dfrac{1}{1000}\right)+\dots\text{ n terms }\right\}

=\dfrac{6}{9}\left\{n-\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\dfrac{1}{10^3}+\dots{\text{ n terms }}\right)\right}

=\frac{6}{9}\left\{n-\frac{1}{10} \frac{\left(1-\left(\frac{1}{10}\right)^{n}\right)}{\left(1-\frac{1}{10}\right)}\right\}

=\frac{6}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right\}

 

QUESTION 5

How many terms of the G.P. 3,3 / 2,3 / 4, \ldots be taken together to make \frac{3069}{512} ?

Sol :

First term , a = 3

Common ratio, r=\frac{1}{2}

S_{n}=\frac{3069}{512}

\therefore S_{n}=a\left(\frac{1-r^{n}}{{1-r}^{\phantom{n}}}\right)

S_{n}=3\left\{\frac{1-\left(\frac{1}{2}\right)^{n}}{1-\frac{1}{2}}\right\}

\Rightarrow \frac{3069}{512}=3\left\{\frac{1-\frac{1}{2^{n}}}{\frac{1}{2}}\right\}

\Rightarrow \frac{3069}{512}=6\left\{1-\frac{1}{2^{n}}\right\}

\Rightarrow \frac{3069}{3072}=1-\frac{1}{2^{n}}

\Rightarrow \frac{1}{2^{n}}=1-\frac{3069}{3072}

\Rightarrow \frac{1}{2^{n}}=\frac{3}{3072}

\Rightarrow 2^{n}=\frac{3072}{3}

\Rightarrow 2^{n}=1024

\Rightarrow 2^{n}=2^{10}

n = 10

 

QUESTION 6

How many terms of the series 2+6+18+\ldots must be taken to make the sum equal to 728 ?

Sol :

First term , a = 2

Common ratio, r = 3

Given : S_{n}=728

S_{n}=a\left(\dfrac{r^n-1}{r-1}\right)

728=2\left(\frac{3^{n}-1}{3-1}\right)

\Rightarrow 728=3^{n}-1

\Rightarrow 3^{n}=729

\Rightarrow 3^{n}=3^{6}

n = 6

 

QUESTION 7

How many terms of the sequence \sqrt{3}, 3,3 \sqrt{3}, \ldots must be taken to make the sum 39+13 \sqrt{3} ?

Sol :

First term , a=\sqrt{3}

Common ratio, r=\sqrt{3}

S_{n}=39+13 \sqrt{3}

S_{n}=a\left(\dfrac{r^n-1}{r-1}\right)

39+13 \sqrt{3}=\sqrt{3}\left(\frac{(\sqrt{3})^{n}-1}{\sqrt{3}-1}\right)

\Rightarrow 39+13 \sqrt{3}=\frac{\sqrt{3}}{(\sqrt{3}-1)}\left\{(\sqrt{3})^{n}-1\right\}

\Rightarrow(\sqrt{3})^{n}-1=\frac{(39+13 \sqrt{3})(\sqrt{3}-1)}{\sqrt{3}}

\Rightarrow(\sqrt{3})^{n}=1+26

\Rightarrow(\sqrt{3})^{n}=27

\Rightarrow(\sqrt{3})^{n}=(\sqrt{3})^{6}

n = 6

 

QUESTION 8

The sum of n terms of the G.P. 3,6,12, \ldots is 381 . Find the value of n

Sol :

First term , a = 3

Common ratio, r = 3

Sum of n terms, S_{n}=381

S_{n}=a\left(\dfrac{r^n-1}{r-1}\right)

\Rightarrow 381=3\left(\frac{2^{n}-1}{2-1}\right)

\Rightarrow 381=3\left(2^{n}-1\right)

\Rightarrow 127=2^{n}-1

\Rightarrow 2^{n}=128

\Rightarrow 2^{n}=2^{7}

n = 7

 

QUESTION 9

The common ratio of a G.P. is 3 and the last term is 486 . If the sum of these terms be 728, find the first term.

Sol :

common ratio, r = 3

n^{\text { th }} term, a_{n}=486

S_{n}=728

a_{n}=486

\Rightarrow a r^{n-1}=486

\Rightarrow a(3)^{n-1}=486

\Rightarrow a(3)^{n}=486 \times 3

\Rightarrow a(3)^{n}=1458~ \ldots (i)

Now, S_{n}=728

\Rightarrow 728=a\left(\frac{3^{n}-1}{3-1}\right)

\Rightarrow 728=\left\{\frac{a(3)^{n}-a}{2}\right\}

\Rightarrow 1456=a(3)^{n-1}-a

\Rightarrow 1456=1458-a [from(i)]

a = 1458 – 1456

a = 2

 

QUESTION 10

The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152 . Find the common ratio.

Sol :

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

\therefore S_{3}=a\left(\frac{r^{3}-1}{r-1}\right) \quad and S_{6}=a\left(\frac{r^{6}-1}{r-1}\right)

Then, according to the question

\frac{S_{3}}{S_{6}}=\frac{a\left(\frac{r^{3}-1}{r-1}\right)}{a\left(\frac{r^{6}-1}{r-1}\right)}

\Rightarrow \frac{125}{152}=\frac{r^{3}-1}{r^{6}-1}

\Rightarrow 125\left(r^{6}-1\right)=152\left(r^{3}-1\right)

\Rightarrow 125 r^{6}-125=152 r^{3}-152

\Rightarrow 125 r^{6}-152 r^{3}+27=0

Now, let r^{3}=y

\therefore 125 y^{2}-152 y+27=0

Now, applying the quadratic formula

y=\left\{\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right\}

\Rightarrow y=\left\{\frac{152 \pm \sqrt{9604}}{250}\right\}

\Rightarrow y=\left\{\frac{152+\sqrt{9604}}{250}\right\} or \left\{\frac{152-\sqrt{9604}}{250}\right\}

\Rightarrow y=1 or \frac{27}{125}

\therefore r^{3}=1 or \quad r^{3}=\frac{27}{125}

But, r=1 is not possible.

\therefore r=\sqrt[3]{\frac{27}{125}}=\frac{3}{5}

 

QUESTION 11

The 4 th and 7 th terms of a G.P. are \frac{1}{27} and \frac{1}{729} respectively. Find the sum of n terms of the G.P.

Sol :

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

 

QUESTION 12

QUESTION 13

QUESTION 14

QUESTION 15

QUESTION 16

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