Question

Find the sum of the following geometric series :

(i) 0.15 + 0.015 + 0.0015 + ..... to 8 terms ;

(ii) $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+.....$ to 8 terms ;

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

(iv) $(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+xy+y^3)+....$ to n terms ;

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

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

(vii) $1,-a,a^2,-a^3,........$ to n terms $(a\ne 1)$

(viii) $x^3,x^5,x^7,....$ to n terms

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

Answer 1

(i) 0.15 + 0.015 + 0.0015 + .... upto 8 terms

Here, a = 0.15 and $r=\frac{a_2}{a_1}=\frac{0.015}{0.15}=\frac{1}{10}$.

$S_8=a\left(\frac{1-r^8}{1-r}\right)$

$=0.15\left(\frac{1-{\left(\frac{1}{10}\right)}^8}{1-\frac{1}{10}}\right)=0.15\left(\frac{1-\frac{1}{{10}^8}}{\frac{1}{10}}\right)=\frac{1}{6}(1-\frac{1}{{10}^8})$

(ii) $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+....$ to 8 terms ;

Here the first term of the series is $a=\sqrt{2}$ and the common ratio is $r=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$

Thus the sum of the G.P. up to 8th terms is :

$S_8=\frac{a(1-r^8)}{1-r}=\frac{\sqrt{2}(1-{\left(\frac{1}{2}\right)}^8)}{1-\frac{1}{2}}$

$=2\sqrt{2}(1-\frac{1}{256})=\frac{255\sqrt{2}}{128}$

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

$a=\frac{2}{9},r=\frac{\frac{-1}{3}}{\frac{2}{9}}=\frac{-1}{3}\times \frac{9}{2}=\frac{-3}{2},n=5$

$S_5=\frac{(1-r^5)}{1-r}$

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

$=\frac{2}{9}\frac{(1+\frac{243}{32})}{1+\frac{3}{2}}$

$=\frac{2}{9}\frac{\left(275\right)}{32}\times \frac{2}{5}=\frac{55}{72}$

(iv)

$(x+y)+(x^2+xy+y^3)+(x^3+x^{2}y+x{y}^2+y^3)+.....$

$=\frac{1}{x-y}\left\{\right(x^2-y^2)+(x^3-y^3)+...to\infty \}..........$

$[\because \frac{x^n-y^n}{x-y}=x^{n-1}+x^{n-2}y+...+y^{n-1}]$

$=\frac{1}{x-y}\left\{\right(x^2+x^3+....to\infty )-(y^2+y^3+....to\infty \left)\right\}$

$=\frac{1}{x-y}\{\frac{x^2}{1-x}-\frac{y^2}{1-y}\}$

$=\frac{1}{x-y}\left\{\frac{x^2-x^{2}y-y^2+x{y}^2}{(1-x)(1-y)}\right\}$

$=\frac{x+y-xy}{(1-x)(1-y)}$

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

The series can be written as :

$3(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+....nterms)+4(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+....nterms)$

For the first part $a=\frac{1}{5}$ and the common ratio $r=\frac{1}{5^2}=\frac{1}{25}$

Thus the sum is :

$3(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...nterms)=3\frac{\frac{1}{5}(1-{\left(\frac{1}{25}\right)}^n)}{1-\frac{1}{25}}$

$=\frac{5}{8}(1-\frac{1}{5^{2n}})$

For the second part $a=\frac{1}{25}$ and common ratio

$r=\frac{1}{25}$ then

$4(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+.....nterms)=4.\frac{\frac{1}{25}(1-{\left(\frac{1}{25}\right)}^n)}{1-\frac{1}{25}}=\frac{1}{6}(1-\frac{1}{5^{2n}})$

Thus the sum is :

$\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^2}+....2n$ terms

$=\frac{5}{8}(1-\frac{1}{5^{2n}})+\frac{1}{6}(1-\frac{1}{5^{2n}})$

$=\frac{5}{8}+\frac{1}{6}(1-\frac{1}{5^{2n}})$

$=\frac{15+4}{24}(1-\frac{1}{5^{2n}})$

$=\frac{19}{24}(1-\frac{1}{5^{2n}})$

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

$a=\frac{a}{1+i},r=\frac{\frac{a}{(1+i{)}^2}}{\frac{a}{1+i}}=\frac{1}{1+i}{S}_n=a\frac{(1-r^n)}{1-r}=\frac{a}{1+i}\frac{(1-{\left(\frac{1}{1+i}\right)}^n)}{1-\frac{1}{1+i}}=\frac{a}{1+i}\times \frac{1+i}{(-i)}(1-(1+i{)}^n)=-ai(1-(1+i{)}^{-n})$

(vii) $1,-a,a^2,-a^3,....$ to n terms $(a\ne 1)$ Rewriting the sequence and sum we get, $Sum=1-a+a^2-a^3+a^4-a^5+....$

Here, r = - a and first term = 1

$Sum=\frac{[1-(-a{)}^n]}{1+a}$

(viii) $x^3,x^5,x^7,.....$ to n terms

Here the first term of the G.P, is a = $x^3$

and the common ratio is $r=\frac{x^5}{x^3}=x^2$

Thus the sum of the G.P. is

$x^3+x^5+x^7+....$ to n terms

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

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

Here the first term of the G.P. is a = $\sqrt{7}$

(ix) $\sqrt{7},\sqrt{21},3\sqrt{7},....$ to terms

Here the first term of the G.P. is a = $\sqrt{7}$

and the common ratio is $r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$

Thus the sum of the G.P. is :

$\sqrt{7}+\sqrt{21}+3\sqrt{7}+....$ to n terms

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