Class 11: Geometric Progression

Notes:

$\displaystyle \text{We know } S_n = a \Big( \frac{r^n - 1}{r-1} \Big) \text{ when } r > 1 \text{and } S_n = a \Big( \frac{1- r^n }{1 - r} \Big) \text{ when } r < 1$
$\displaystyle a_n = ar^{n-1}$

$\displaystyle \textbf{Question 1: }\text{Find the sum of the following geometric progressions:}$
$\displaystyle \text{(i) }2,6,18,\ldots\text{ to }7\text{ terms}$
$\displaystyle \text{(ii) }1,3,9,27,\ldots\text{ to }8\text{ terms}$
$\displaystyle \text{(iii) }1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8},\ldots\text{ to }9\text{ terms}$
$\displaystyle \text{(iv) }(a^2-b^2),(a-b),\frac{a-b}{a+b},\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(v) }4,2,1,\frac{1}{2},\ldots\text{ to }10\text{ terms}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{We use }S_n=\frac{a(r^n-1)}{r-1}\text{ when }r>1,$
$\displaystyle \text{and }S_n=\frac{a(1-r^n)}{1-r}\text{ when }r<1.$
$\displaystyle \text{(i) Given G.P.: }2,6,18,\ldots$
$\displaystyle \text{Here, }a=2,\qquad r=\frac{6}{2}=3,\qquad n=7.$
$\displaystyle S_7=\frac{2(3^7-1)}{3-1}$
$\displaystyle =3^7-1=2187-1=2186.$
$\displaystyle \therefore \text{The required sum is }2186.$
$\displaystyle \text{(ii) Given G.P.: }1,3,9,27,\ldots$
$\displaystyle \text{Here, }a=1,\qquad r=\frac{3}{1}=3,\qquad n=8.$
$\displaystyle S_8=\frac{1(3^8-1)}{3-1}$
$\displaystyle =\frac{6561-1}{2}=\frac{6560}{2}=3280.$
$\displaystyle \therefore \text{The required sum is }3280.$
$\displaystyle \text{(iii) Given G.P.: }1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8},\ldots$
$\displaystyle \text{Here, }a=1,\qquad r=-\frac{1}{2},\qquad n=9.$
$\displaystyle S_9=\frac{1-\left(-\frac{1}{2}\right)^9}{1-\left(-\frac{1}{2}\right)}$
$\displaystyle =\frac{1+\frac{1}{512}}{\frac{3}{2}}$
$\displaystyle =\frac{513}{512}\times\frac{2}{3}=\frac{171}{256}.$
$\displaystyle \therefore \text{The required sum is }\frac{171}{256}.$
$\displaystyle \text{(iv) Given G.P.: }(a^2-b^2),(a-b),\frac{a-b}{a+b},\ldots$
$\displaystyle \text{Let the first term be }A=a^2-b^2.$
$\displaystyle r=\frac{a-b}{a^2-b^2}$
$\displaystyle =\frac{a-b}{(a-b)(a+b)}=\frac{1}{a+b},\qquad a+b\neq0.$
$\displaystyle S_n=(a^2-b^2)\left[\frac{1-\left(\frac{1}{a+b}\right)^n}{1-\frac{1}{a+b}}\right]$
$\displaystyle =(a-b)(a+b)\left[\frac{1-\frac{1}{(a+b)^n}}{\frac{a+b-1}{a+b}}\right]$
$\displaystyle =\frac{(a-b)(a+b)^2}{a+b-1}\left[\frac{(a+b)^n-1}{(a+b)^n}\right]$
$\displaystyle =\frac{(a-b)\left[(a+b)^n-1\right]}{(a+b)^{n-2}(a+b-1)}.$
$\displaystyle \therefore S_n=\frac{(a-b)\left[(a+b)^n-1\right]}{(a+b)^{n-2}(a+b-1)},\qquad a+b\neq1.$
$\displaystyle \text{If }a+b=1,\text{ then }r=1\text{ and }S_n=n(a^2-b^2)=n(a-b).$
$\displaystyle \text{(v) Given G.P.: }4,2,1,\frac{1}{2},\ldots$
$\displaystyle \text{Here, }a=4,\qquad r=\frac{2}{4}=\frac{1}{2},\qquad n=10.$
$\displaystyle S_{10}=\frac{4\left[1-\left(\frac{1}{2}\right)^{10}\right]}{1-\frac{1}{2}}$
$\displaystyle =8\left(1-\frac{1}{1024}\right)$
$\displaystyle =8\times\frac{1023}{1024}=\frac{1023}{128}.$
$\displaystyle \therefore \text{The required sum is }\frac{1023}{128}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the sum of the following geometric progressions:}$
$\displaystyle \text{(i) }0.15+0.015+0.0015+\ldots\text{ to }8\text{ terms}$
$\displaystyle \text{(ii) }\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\ldots\text{ to }8\text{ terms}$
$\displaystyle \text{(iii) }\frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+\ldots\text{ to }5\text{ terms}$
$\displaystyle \text{(iv) }(x+y)+(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+\ldots$
$\displaystyle \text{to }n\text{ terms}$
$\displaystyle \text{(v) }\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+\ldots\text{ to }2n\text{ terms}$
$\displaystyle \text{(vi) }\frac{a}{1+i}+\frac{a}{(1+i)^2}+\frac{a}{(1+i)^3}+\ldots+\frac{a}{(1+i)^n}$
$\displaystyle \text{(vii) }1,-a,a^2,-a^3,\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(viii) }x^3,x^5,x^7,\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(ix) }\sqrt{7},\sqrt{21},3\sqrt{7},\ldots\text{ to }n\text{ terms}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{(i) Given G.P.: }0.15+0.015+0.0015+\ldots$
$\displaystyle \text{Here, }a=0.15,\qquad r=\frac{0.015}{0.15}=0.1,\qquad n=8.$
$\displaystyle S_8=\frac{0.15(1-0.1^8)}{1-0.1}$
$\displaystyle =\frac{0.15}{0.9}(1-0.1^8)$
$\displaystyle =\frac{1}{6}\left(1-\frac{1}{10^8}\right).$
$\displaystyle \therefore \text{The required sum is }\frac{1}{6}\left(1-\frac{1}{10^8}\right).$
$\displaystyle \text{(ii) Given G.P.: }\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\ldots$
$\displaystyle \text{Here, }a=\sqrt{2},\qquad r=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2},\qquad n=8.$
$\displaystyle S_8=\frac{\sqrt{2}\left[1-\left(\frac{1}{2}\right)^8\right]}{1-\frac{1}{2}}$
$\displaystyle =2\sqrt{2}\left(1-\frac{1}{256}\right)$
$\displaystyle =2\sqrt{2}\times\frac{255}{256}=\frac{255\sqrt{2}}{128}.$
$\displaystyle \therefore \text{The required sum is }\frac{255\sqrt{2}}{128}.$
$\displaystyle \text{(iii) Given G.P.: }\frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+\ldots$
$\displaystyle \text{Here, }a=\frac{2}{9},\qquad r=\frac{-\frac{1}{3}}{\frac{2}{9}}=-\frac{3}{2},\qquad n=5.$
$\displaystyle S_5=\frac{\frac{2}{9}\left[1-\left(-\frac{3}{2}\right)^5\right]}{1+\frac{3}{2}}$
$\displaystyle =\frac{2}{9}\left[\frac{1+\frac{243}{32}}{\frac{5}{2}}\right]$
$\displaystyle =\frac{2}{9}\times\frac{275}{32}\times\frac{2}{5}$
$\displaystyle =\frac{55}{72}.$
$\displaystyle \therefore \text{The required sum is }\frac{55}{72}.$
$\displaystyle \text{(iv) Let the required sum be }S_n.$
$\displaystyle S_n=(x+y)+(x^2+xy+y^2)+\ldots$
$\displaystyle \qquad +(x^n+x^{n-1}y+\ldots+xy^{n-1}+y^n).$
$\displaystyle \text{Using }\frac{x^{m+1}-y^{m+1}}{x-y}=x^m+x^{m-1}y+\ldots+y^m,$
$\displaystyle S_n=\frac{1}{x-y}\left[(x^2-y^2)+(x^3-y^3)+\ldots+(x^{n+1}-y^{n+1})\right].$
$\displaystyle S_n=\frac{1}{x-y}\left[(x^2+x^3+\ldots+x^{n+1})-(y^2+y^3+\ldots+y^{n+1})\right].$
$\displaystyle S_n=\frac{1}{x-y}\left[\frac{x^2(x^n-1)}{x-1}-\frac{y^2(y^n-1)}{y-1}\right],$
$\displaystyle \text{where }x\neq y,\quad x\neq1\quad\text{and}\quad y\neq1.$
$\displaystyle \text{(v) Grouping the }2n\text{ terms in }n\text{ pairs,}$
$\displaystyle S_{2n}=\left(\frac{3}{5}+\frac{4}{5^2}\right)\left(1+\frac{1}{5^2}+\frac{1}{5^4}+\ldots+\frac{1}{5^{2n-2}}\right).$
$\displaystyle =\frac{19}{25}\left[\frac{1-\left(\frac{1}{25}\right)^n}{1-\frac{1}{25}}\right]$
$\displaystyle =\frac{19}{25}\times\frac{25}{24}\left(1-\frac{1}{5^{2n}}\right)$
$\displaystyle =\frac{19}{24}\left(1-\frac{1}{5^{2n}}\right).$
$\displaystyle \therefore \text{The required sum is }\frac{19}{24}\left(1-\frac{1}{5^{2n}}\right).$
$\displaystyle \text{(vi) Given G.P.: }\frac{a}{1+i}+\frac{a}{(1+i)^2}+\ldots+\frac{a}{(1+i)^n}.$
$\displaystyle \text{Here, the first term }A=\frac{a}{1+i},\qquad r=\frac{1}{1+i}.$
$\displaystyle S_n=\frac{a}{1+i}\left[\frac{1-\left(\frac{1}{1+i}\right)^n}{1-\frac{1}{1+i}}\right]$
$\displaystyle =\frac{a}{1+i}\left[\frac{1-(1+i)^{-n}}{\frac{i}{1+i}}\right]$
$\displaystyle =\frac{a}{i}\left[1-(1+i)^{-n}\right]$
$\displaystyle =-ai\left[1-(1+i)^{-n}\right].$
$\displaystyle \therefore \text{The required sum is }-ai\left[1-(1+i)^{-n}\right].$
$\displaystyle \text{(vii) Given G.P.: }1,-a,a^2,-a^3,\ldots$
$\displaystyle \text{Here, the first term }A=1,\qquad r=-a.$
$\displaystyle S_n=\frac{1-(-a)^n}{1-(-a)}$
$\displaystyle =\frac{1-(-a)^n}{1+a},\qquad a\neq-1.$
$\displaystyle \text{If }a=-1,\text{ then every term is }1\text{ and }S_n=n.$
$\displaystyle \text{(viii) Given G.P.: }x^3,x^5,x^7,\ldots$
$\displaystyle \text{Here, }a=x^3,\qquad r=\frac{x^5}{x^3}=x^2.$
$\displaystyle S_n=\frac{x^3(1-x^{2n})}{1-x^2}$
$\displaystyle =\frac{x^3(x^{2n}-1)}{x^2-1},\qquad x\neq\pm1.$
$\displaystyle \text{If }x=1,\text{ then }S_n=n;\text{ and if }x=-1,\text{ then }S_n=-n.$
$\displaystyle \text{(ix) Given G.P.: }\sqrt{7},\sqrt{21},3\sqrt{7},\ldots$
$\displaystyle \text{Here, }a=\sqrt{7},\qquad r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}.$
$\displaystyle S_n=\frac{\sqrt{7}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1}.$
$\displaystyle \therefore \text{The required sum is }\frac{\sqrt{7}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }\sum\limits_{n=1}^{11}(2+3^n)\qquad\text{(ii) }\sum\limits_{k=1}^{n}(2^k+3^{k-1})$
$\displaystyle \text{(iii) }\sum\limits_{n=2}^{10}4^n$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{(i) }\sum\limits_{n=1}^{11}(2+3^n)$
$\displaystyle =(2+3^1)+(2+3^2)+(2+3^3)+\ldots+(2+3^{11})$
$\displaystyle =2\times11+(3^1+3^2+3^3+\ldots+3^{11})$
$\displaystyle =22+3(1+3+3^2+\ldots+3^{10})$
$\displaystyle =22+3\left(\frac{3^{11}-1}{3-1}\right)$
$\displaystyle =22+\frac{3}{2}(3^{11}-1)$
$\displaystyle =22+\frac{3}{2}(177147-1)$
$\displaystyle =22+\frac{3}{2}\times177146$
$\displaystyle =22+265719=265741.$
$\displaystyle \therefore \sum\limits_{n=1}^{11}(2+3^n)=265741.$
$\displaystyle \text{(ii) }\sum\limits_{k=1}^{n}(2^k+3^{k-1})$
$\displaystyle =(2^1+3^0)+(2^2+3^1)+(2^3+3^2)+\ldots+(2^n+3^{n-1})$
$\displaystyle =(2^1+2^2+2^3+\ldots+2^n)+(3^0+3^1+3^2+\ldots+3^{n-1})$
$\displaystyle =2\left(\frac{2^n-1}{2-1}\right)+\left(\frac{3^n-1}{3-1}\right)$
$\displaystyle =2(2^n-1)+\frac{1}{2}(3^n-1)$
$\displaystyle =2^{n+1}-2+\frac{3^n-1}{2}$
$\displaystyle =\frac{2^{n+2}+3^n-5}{2}.$
$\displaystyle \therefore \sum\limits_{k=1}^{n}(2^k+3^{k-1})=\frac{2^{n+2}+3^n-5}{2}.$
$\displaystyle \text{(iii) }\sum\limits_{n=2}^{10}4^n$
$\displaystyle =4^2+4^3+\ldots+4^{10}$
$\displaystyle =4^2(1+4+4^2+\ldots+4^8)$
$\displaystyle =4^2\left(\frac{4^9-1}{4-1}\right)$
$\displaystyle =\frac{16}{3}(4^9-1)$
$\displaystyle =\frac{16}{3}(262144-1)$
$\displaystyle =\frac{16}{3}\times262143$
$\displaystyle =16\times87381=1398096.$
$\displaystyle \therefore \sum\limits_{n=2}^{10}4^n=1398096.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the sum of the following series:}$
$\displaystyle \text{(i) }5+55+555+\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(ii) }7+77+777+\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(iii) }9+99+999+\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(iv) }0.5+0.55+0.555+\ldots\text{ to }n\text{ terms}$
$\displaystyle \text{(v) }0.6+0.66+0.666+\ldots\text{ to }n\text{ terms}$
$\displaystyle \textbf{Answer:}$

$\displaystyle \text{(i) Let }S_n=5+55+555+\ldots\text{ to }n\text{ terms}.$
$\displaystyle S_n=5(1+11+111+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{5}{9}(9+99+999+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{5}{9}\left[(10-1)+(10^2-1)+(10^3-1)+\ldots+(10^n-1)\right]$
$\displaystyle =\frac{5}{9}\left[(10+10^2+10^3+\ldots+10^n)-n\right]$
$\displaystyle =\frac{5}{9}\left[\frac{10(10^n-1)}{10-1}-n\right]$
$\displaystyle =\frac{5}{9}\left[\frac{10}{9}(10^n-1)-n\right]$
$\displaystyle =\frac{5}{81}\left(10^{n+1}-9n-10\right).$
$\displaystyle \therefore \text{The required sum is }\frac{5}{81}\left(10^{n+1}-9n-10\right).$

$\displaystyle \text{(ii) Let }S_n=7+77+777+\ldots\text{ to }n\text{ terms}.$
$\displaystyle S_n=7(1+11+111+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{7}{9}(9+99+999+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{7}{9}\left[(10-1)+(10^2-1)+(10^3-1)+\ldots+(10^n-1)\right]$
$\displaystyle =\frac{7}{9}\left[(10+10^2+10^3+\ldots+10^n)-n\right]$
$\displaystyle =\frac{7}{9}\left[\frac{10(10^n-1)}{10-1}-n\right]$
$\displaystyle =\frac{7}{9}\left[\frac{10}{9}(10^n-1)-n\right]$
$\displaystyle =\frac{7}{81}\left(10^{n+1}-9n-10\right).$
$\displaystyle \therefore \text{The required sum is }\frac{7}{81}\left(10^{n+1}-9n-10\right).$

$\displaystyle \text{(iii) Let }S_n=9+99+999+\ldots\text{ to }n\text{ terms}.$
$\displaystyle S_n=(10-1)+(10^2-1)+(10^3-1)+\ldots+(10^n-1)$
$\displaystyle =(10+10^2+10^3+\ldots+10^n)-n$
$\displaystyle =\frac{10(10^n-1)}{10-1}-n$
$\displaystyle =\frac{10}{9}(10^n-1)-n$
$\displaystyle =\frac{1}{9}\left(10^{n+1}-9n-10\right).$
$\displaystyle \therefore \text{The required sum is }\frac{1}{9}\left(10^{n+1}-9n-10\right).$

$\displaystyle \text{(iv) Let }S_n=0.5+0.55+0.555+\ldots\text{ to }n\text{ terms}.$
$\displaystyle S_n=5(0.1+0.11+0.111+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{5}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots\text{ to }n\text{ terms}\right]$
$\displaystyle =\frac{5}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\ldots+\left(1-\frac{1}{10^n}\right)\right]$
$\displaystyle =\frac{5}{9}\left[n-\left(\frac{1}{10}+\frac{1}{10^2}+\ldots+\frac{1}{10^n}\right)\right]$
$\displaystyle =\frac{5}{9}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^n}\right)}{1-\frac{1}{10}}\right]$
$\displaystyle =\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right].$
$\displaystyle \therefore \text{The required sum is }\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right].$

$\displaystyle \text{(v) Let }S_n=0.6+0.66+0.666+\ldots\text{ to }n\text{ terms}.$
$\displaystyle S_n=6(0.1+0.11+0.111+\ldots\text{ to }n\text{ terms})$
$\displaystyle =\frac{6}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots\text{ to }n\text{ terms}\right]$
$\displaystyle =\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\ldots+\left(1-\frac{1}{10^n}\right)\right]$
$\displaystyle =\frac{6}{9}\left[n-\left(\frac{1}{10}+\frac{1}{10^2}+\ldots+\frac{1}{10^n}\right)\right]$
$\displaystyle =\frac{6}{9}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^n}\right)}{1-\frac{1}{10}}\right]$
$\displaystyle =\frac{6}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right].$
$\displaystyle \therefore \text{The required sum is }\frac{2}{3}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right].$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{How many terms of the G.P. }3,\frac{3}{2},\frac{3}{4},\ldots\text{ must be}$
$\displaystyle \text{taken together to make the sum }\frac{3069}{512}\text{?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }a=3,\qquad r=\frac{\frac{3}{2}}{3}=\frac{1}{2},\qquad S_n=\frac{3069}{512}.$
$\displaystyle S_n=\frac{a(1-r^n)}{1-r}$
$\displaystyle \frac{3069}{512}=\frac{3\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}}$
$\displaystyle \frac{3069}{512}=6\left[1-\left(\frac{1}{2}\right)^n\right]$
$\displaystyle \frac{3069}{3072}=1-\left(\frac{1}{2}\right)^n$
$\displaystyle \left(\frac{1}{2}\right)^n=1-\frac{3069}{3072}$
$\displaystyle =\frac{3}{3072}=\frac{1}{1024}=\frac{1}{2^{10}}$
$\displaystyle \therefore n=10.$
$\displaystyle \therefore \text{The required number of terms is }10.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{How many terms of the series }2+6+18+\ldots\text{ must be taken to make}$
$\displaystyle \text{the sum equal to }728\text{?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }a=2,\qquad r=\frac{6}{2}=3,\qquad S_n=728.$
$\displaystyle S_n=\frac{a(r^n-1)}{r-1}$
$\displaystyle 728=\frac{2(3^n-1)}{3-1}$
$\displaystyle 728=3^n-1$
$\displaystyle 3^n=729$
$\displaystyle 3^n=3^6$
$\displaystyle \therefore n=6.$
$\displaystyle \therefore \text{The required number of terms is }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{How many terms of the sequence }\sqrt{3},3,3\sqrt{3},\ldots\text{ must be taken}$
$\displaystyle \text{to make the sum }39+13\sqrt{3}\text{?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }a=\sqrt{3},\qquad r=\frac{3}{\sqrt{3}}=\sqrt{3},\qquad S_n=39+13\sqrt{3}.$
$\displaystyle S_n=\frac{a(r^n-1)}{r-1}$
$\displaystyle 39+13\sqrt{3}=\frac{\sqrt{3}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1}$
$\displaystyle (\sqrt{3})^n-1=\frac{(39+13\sqrt{3})(\sqrt{3}-1)}{\sqrt{3}}$
$\displaystyle =\frac{39\sqrt{3}-39+39-13\sqrt{3}}{\sqrt{3}}$
$\displaystyle =\frac{26\sqrt{3}}{\sqrt{3}}=26.$
$\displaystyle (\sqrt{3})^n=27$
$\displaystyle (\sqrt{3})^n=(\sqrt{3})^6$
$\displaystyle \therefore n=6.$
$\displaystyle \therefore \text{The required number of terms is }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The sum of }n\text{ terms of the G.P. }3,6,12,\ldots\text{ is }381.\text{ Find the value}$
$\displaystyle \text{of }n\text{.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }a=3,\qquad r=\frac{6}{3}=2,\qquad S_n=381.$
$\displaystyle S_n=\frac{a(r^n-1)}{r-1}$
$\displaystyle 381=\frac{3(2^n-1)}{2-1}$
$\displaystyle 381=3(2^n-1)$
$\displaystyle 127=2^n-1$
$\displaystyle 2^n=128=2^7$
$\displaystyle \therefore n=7.$
$\displaystyle \therefore \text{The required number of terms is }7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The common ratio of a G.P. is }3\text{ and the last term is }486.\text{ If the sum}$
$\displaystyle \text{of its terms is }728,\text{ find the first term.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }r=3,\qquad a_n=486,\qquad S_n=728.$
$\displaystyle \text{We know, }a_n=ar^{n-1}.$
$\displaystyle 486=a\cdot3^{n-1}\qquad\ldots\ldots\text{(i)}$
$\displaystyle \text{Also, }S_n=\frac{a(r^n-1)}{r-1}.$
$\displaystyle 728=\frac{a(3^n-1)}{3-1}$
$\displaystyle 1456=a(3^n-1)\qquad\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \frac{1456}{486}=\frac{3^n-1}{3^{n-1}}$
$\displaystyle \frac{1456}{486}=3-\frac{1}{3^{n-1}}$
$\displaystyle \frac{1}{3^{n-1}}=3-\frac{1456}{486}$
$\displaystyle =\frac{1458-1456}{486}=\frac{2}{486}$
$\displaystyle =\frac{1}{243}=\frac{1}{3^5}.$
$\displaystyle 3^{n-1}=3^5$
$\displaystyle n-1=5$
$\displaystyle n=6.$
$\displaystyle \text{Substituting }n=6\text{ in (i),}$
$\displaystyle 486=a\cdot3^5$
$\displaystyle 486=243a$
$\displaystyle a=2.$
$\displaystyle \therefore \text{The first term of the G.P. is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The ratio of the sum of the first three terms to the sum of the first six terms}$
$\displaystyle \text{of a G.P. is }125:152.\text{ Find the common ratio.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Given, }\frac{S_3}{S_6}=\frac{125}{152}.$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle S_3=a(1+r+r^2).$
$\displaystyle S_6=a(1+r+r^2+r^3+r^4+r^5)$
$\displaystyle =a(1+r+r^2)(1+r^3)$
$\displaystyle =S_3(1+r^3).$
$\displaystyle \therefore \frac{S_3}{S_6}=\frac{1}{1+r^3}.$
$\displaystyle \frac{1}{1+r^3}=\frac{125}{152}$
$\displaystyle 152=125(1+r^3)$
$\displaystyle 152=125+125r^3$
$\displaystyle 125r^3=27$
$\displaystyle r^3=\frac{27}{125}$
$\displaystyle r^3=\left(\frac{3}{5}\right)^3$
$\displaystyle \therefore r=\frac{3}{5}.$
$\displaystyle \therefore \text{The common ratio is }\frac{3}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The fourth and the seventh terms of a G.P. are }\frac{1}{27}\text{ and }\frac{1}{729}$
$\displaystyle \text{respectively. Find the sum of }n\text{ terms of the G.P.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Given, }a_4=\frac{1}{27}\qquad\text{and}\qquad a_7=\frac{1}{729}.$
$\displaystyle ar^3=\frac{1}{27}\qquad\ldots\ldots\text{(i)}$
$\displaystyle ar^6=\frac{1}{729}\qquad\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \frac{ar^6}{ar^3}=\frac{\frac{1}{729}}{\frac{1}{27}}$
$\displaystyle r^3=\frac{27}{729}=\frac{1}{27}$
$\displaystyle r^3=\left(\frac{1}{3}\right)^3$
$\displaystyle \therefore r=\frac{1}{3}.$
$\displaystyle \text{Substituting }r=\frac{1}{3}\text{ in (i),}$
$\displaystyle a\left(\frac{1}{3}\right)^3=\frac{1}{27}$
$\displaystyle \frac{a}{27}=\frac{1}{27}$
$\displaystyle \therefore a=1.$
$\displaystyle S_n=\frac{a(1-r^n)}{1-r}$
$\displaystyle =\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}$
$\displaystyle =\frac{3}{2}\left(1-\frac{1}{3^n}\right).$
$\displaystyle \therefore \text{The sum of }n\text{ terms is }\frac{3}{2}\left(1-\frac{1}{3^n}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Find the sum }\sum\limits_{n=1}^{10}\left\{\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right\}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \sum\limits_{n=1}^{10}\left\{\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right\}$
$\displaystyle =\left[\left(\frac{1}{2}\right)^0+\left(\frac{1}{5}\right)^2\right]+\left[\left(\frac{1}{2}\right)^1+\left(\frac{1}{5}\right)^3\right]+\ldots$
$\displaystyle \qquad+\left[\left(\frac{1}{2}\right)^9+\left(\frac{1}{5}\right)^{11}\right]$
$\displaystyle =\left(\frac{1}{2^0}+\frac{1}{2^1}+\ldots+\frac{1}{2^9}\right)+\left(\frac{1}{5^2}+\frac{1}{5^3}+\ldots+\frac{1}{5^{11}}\right)$
$\displaystyle =\frac{1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}}+\frac{1}{5^2}\left[\frac{1-\left(\frac{1}{5}\right)^{10}}{1-\frac{1}{5}}\right]$
$\displaystyle =2\left(1-\frac{1}{2^{10}}\right)+\frac{1}{20}\left(1-\frac{1}{5^{10}}\right)$
$\displaystyle =\frac{2^{10}-1}{2^9}+\frac{5^{10}-1}{4\cdot5^{11}}.$
$\displaystyle \therefore \text{The required sum is }\frac{2^{10}-1}{2^9}+\frac{5^{10}-1}{4\cdot5^{11}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The fifth term of a G.P. is }81,\text{ whereas its second term is }24.$
$\displaystyle \text{Find the G.P. and the sum of its first eight terms.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle \text{Given, }a_2=24.$
$\displaystyle ar=24\qquad\ldots\ldots\text{(i)}$
$\displaystyle \text{Also, }a_5=81.$
$\displaystyle ar^4=81\qquad\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \frac{ar^4}{ar}=\frac{81}{24}$
$\displaystyle r^3=\frac{27}{8}$
$\displaystyle r^3=\left(\frac{3}{2}\right)^3$
$\displaystyle \therefore r=\frac{3}{2}.$
$\displaystyle \text{Substituting }r=\frac{3}{2}\text{ in (i),}$
$\displaystyle a\left(\frac{3}{2}\right)=24$
$\displaystyle a=24\times\frac{2}{3}=16.$
$\displaystyle \therefore \text{The G.P. is }16,24,36,54,81,\frac{243}{2},\frac{729}{4},\frac{2187}{8},\ldots$
$\displaystyle S_8=\frac{a(r^8-1)}{r-1}$
$\displaystyle =\frac{16\left[\left(\frac{3}{2}\right)^8-1\right]}{\frac{3}{2}-1}$
$\displaystyle =32\left(\frac{3^8}{2^8}-1\right)$
$\displaystyle =32\left(\frac{6561-256}{256}\right)$
$\displaystyle =32\times\frac{6305}{256}$
$\displaystyle =\frac{6305}{8}.$
$\displaystyle \therefore \text{The sum of the first eight terms is }\frac{6305}{8}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }S_1,S_2,S_3\text{ are respectively the sums of }n,2n\text{ and }3n\text{ terms}$
$\displaystyle \text{of a G.P., prove that }S_1^2+S_2^2=S_1(S_2+S_3).$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle \text{First, let }r\neq1.$
$\displaystyle S_1=\frac{a(r^n-1)}{r-1}.$
$\displaystyle S_2=\frac{a(r^{2n}-1)}{r-1}$
$\displaystyle =\frac{a(r^n-1)(r^n+1)}{r-1}$
$\displaystyle =S_1(r^n+1).$
$\displaystyle S_3=\frac{a(r^{3n}-1)}{r-1}$
$\displaystyle =\frac{a(r^n-1)(r^{2n}+r^n+1)}{r-1}$
$\displaystyle =S_1(r^{2n}+r^n+1).$
$\displaystyle \text{Now, LHS}=S_1^2+S_2^2$
$\displaystyle =S_1^2+S_1^2(r^n+1)^2$
$\displaystyle =S_1^2\left[1+r^{2n}+2r^n+1\right]$
$\displaystyle =S_1^2(r^{2n}+2r^n+2)$
$\displaystyle =S_1\left[S_1(r^n+1)+S_1(r^{2n}+r^n+1)\right]$
$\displaystyle =S_1(S_2+S_3)$
$\displaystyle =\text{RHS}.$
$\displaystyle \text{Hence, }S_1^2+S_2^2=S_1(S_2+S_3)\text{ for }r\neq1.$
$\displaystyle \text{If }r=1,\text{ then }S_1=na,\quad S_2=2na,\quad S_3=3na.$
$\displaystyle S_1^2+S_2^2=(na)^2+(2na)^2=5n^2a^2.$
$\displaystyle S_1(S_2+S_3)=na(2na+3na)=5n^2a^2.$
$\displaystyle \therefore S_1^2+S_2^2=S_1(S_2+S_3)\text{ for }r=1\text{ also.}$
$\displaystyle \therefore S_1^2+S_2^2=S_1(S_2+S_3).\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Show that the ratio of the sum of the first }n\text{ terms of a G.P. to the sum}$
$\displaystyle \text{of the terms from the }(n+1)^{\mathrm{th}}\text{ to the }(2n)^{\mathrm{th}}\text{ term is }\frac{1}{r^n}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle \text{First, let }r\neq1.$
$\displaystyle \text{Sum of the first }n\text{ terms is}$
$\displaystyle S_n=\frac{a(r^n-1)}{r-1}.\qquad\ldots\ldots\text{(i)}$
$\displaystyle \text{The first term of the next }n\text{ terms is }a_{n+1}=ar^n.$
$\displaystyle \text{Therefore, the sum of the terms from the }(n+1)^{\mathrm{th}}\text{ to the }(2n)^{\mathrm{th}}\text{ term is}$
$\displaystyle S=ar^n\left(\frac{r^n-1}{r-1}\right).\qquad\ldots\ldots\text{(ii)}$
$\displaystyle \therefore \frac{\text{Sum of the first }n\text{ terms}}{\text{Sum of the next }n\text{ terms}}$
$\displaystyle =\frac{\frac{a(r^n-1)}{r-1}}{ar^n\left(\frac{r^n-1}{r-1}\right)}$
$\displaystyle =\frac{1}{r^n}.$
$\displaystyle \text{If }r=1,\text{ then each term of the G.P. is }a.$
$\displaystyle \therefore \frac{\text{Sum of the first }n\text{ terms}}{\text{Sum of the next }n\text{ terms}}$
$\displaystyle =\frac{na}{na}=1=\frac{1}{1^n}=\frac{1}{r^n}.$
$\displaystyle \therefore \text{The required ratio is }\frac{1}{r^n}.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If }a,b\text{ are the roots of }x^2-3x+p=0\text{ and }c,d\text{ are the roots}$
$\displaystyle \text{of }x^2-12x+q=0,\text{ where }a,b,c,d\text{ form a G.P., prove that}$
$\displaystyle (q+p):(q-p)=17:15.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Since }a,b\text{ are the roots of }x^2-3x+p=0,$
$\displaystyle a+b=3\qquad\text{and}\qquad ab=p.\qquad\ldots\ldots\text{(i)}$
$\displaystyle \text{Since }c,d\text{ are the roots of }x^2-12x+q=0,$
$\displaystyle c+d=12\qquad\text{and}\qquad cd=q.\qquad\ldots\ldots\text{(ii)}$
$\displaystyle \text{Let the common ratio of the G.P. be }r.$
$\displaystyle \therefore b=ar,\qquad c=ar^2,\qquad d=ar^3.$
$\displaystyle \text{Using }a+b=3,$
$\displaystyle a+ar=3$
$\displaystyle a(1+r)=3.\qquad\ldots\ldots\text{(iii)}$
$\displaystyle \text{Using }c+d=12,$
$\displaystyle ar^2+ar^3=12$
$\displaystyle ar^2(1+r)=12.\qquad\ldots\ldots\text{(iv)}$
$\displaystyle \text{Dividing (iv) by (iii), we get}$
$\displaystyle r^2=\frac{12}{3}=4.$
$\displaystyle \therefore r^4=16.$
$\displaystyle \text{Now, }p=ab=a(ar)=a^2r.$
$\displaystyle \text{Also, }q=cd=(ar^2)(ar^3)=a^2r^5.$
$\displaystyle \therefore \frac{q}{p}=\frac{a^2r^5}{a^2r}=r^4=16.$
$\displaystyle \therefore q=16p.$
$\displaystyle \therefore \frac{q+p}{q-p}=\frac{16p+p}{16p-p}$
$\displaystyle =\frac{17p}{15p}=\frac{17}{15}.$
$\displaystyle \therefore (q+p):(q-p)=17:15.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{How many terms of the G.P. }3,\frac{3}{2},\frac{3}{4},\ldots\text{ are needed to give}$
$\displaystyle \text{the sum }\frac{3069}{512}\text{?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Here, }a=3,\qquad r=\frac{\frac{3}{2}}{3}=\frac{1}{2},\qquad S_n=\frac{3069}{512}.$
$\displaystyle S_n=\frac{a(1-r^n)}{1-r}$
$\displaystyle \frac{3069}{512}=\frac{3\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}}$
$\displaystyle \frac{3069}{512}=6\left(1-\frac{1}{2^n}\right)$
$\displaystyle \frac{3069}{3072}=1-\frac{1}{2^n}$
$\displaystyle \frac{1}{2^n}=1-\frac{3069}{3072}$
$\displaystyle =\frac{3}{3072}=\frac{1}{1024}=\frac{1}{2^{10}}.$
$\displaystyle \therefore n=10.$
$\displaystyle \therefore \text{The required number of terms is }10.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{A person has }2\text{ parents, }4\text{ grandparents, }8\text{ great-grandparents and so on.}$
$\displaystyle \text{Find the number of ancestors in the }10\text{ generations preceding his own.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The numbers of ancestors in successive generations form the G.P.}$
$\displaystyle 2,4,8,\ldots$
$\displaystyle \text{Here, }a=2,\qquad r=\frac{4}{2}=2,\qquad n=10.$
$\displaystyle S_{10}=\frac{a(r^{10}-1)}{r-1}$
$\displaystyle =\frac{2(2^{10}-1)}{2-1}$
$\displaystyle =2(2^{10}-1)$
$\displaystyle =2(1024-1)$
$\displaystyle =2\times1023=2046.$
$\displaystyle \therefore \text{The number of ancestors in the preceding }10\text{ generations is }2046.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If }S_1,S_2,\ldots,S_n\text{ are the sums of }n\text{ terms of }n\text{ G.P.s, each}$
$\displaystyle \text{having first term }1\text{ and common ratios }1,2,3,\ldots,n\text{ respectively, prove that}$
$\displaystyle S_1+S_2+2S_3+3S_4+\ldots+(n-1)S_n=1^n+2^n+3^n+\ldots+n^n.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{For }S_1,\text{ the first term is }1\text{ and the common ratio is }1.$
$\displaystyle \therefore S_1=1+1+1+\ldots+1=n.$
$\displaystyle \text{For the second G.P.,}$
$\displaystyle S_2=\frac{2^n-1}{2-1}=2^n-1.$
$\displaystyle \text{Similarly,}$
$\displaystyle S_3=\frac{3^n-1}{3-1}=\frac{3^n-1}{2},$
$\displaystyle S_4=\frac{4^n-1}{4-1}=\frac{4^n-1}{3},$
$\displaystyle \ldots$
$\displaystyle S_n=\frac{n^n-1}{n-1}.$
$\displaystyle \text{Now, LHS}=S_1+S_2+2S_3+3S_4+\ldots+(n-1)S_n$
$\displaystyle =n+(2^n-1)+2\left(\frac{3^n-1}{2}\right)+3\left(\frac{4^n-1}{3}\right)+\ldots$
$\displaystyle \qquad +(n-1)\left(\frac{n^n-1}{n-1}\right)$
$\displaystyle =n+(2^n-1)+(3^n-1)+(4^n-1)+\ldots+(n^n-1)$
$\displaystyle =n+(2^n+3^n+4^n+\ldots+n^n)-(n-1)$
$\displaystyle =1+2^n+3^n+4^n+\ldots+n^n$
$\displaystyle =1^n+2^n+3^n+4^n+\ldots+n^n$
$\displaystyle =\text{RHS}.$
$\displaystyle \therefore S_1+S_2+2S_3+3S_4+\ldots+(n-1)S_n=1^n+2^n+\ldots+n^n.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{A G.P. consists of an even number of terms. If the sum of all the terms is}$
$\displaystyle 5\text{ times the sum of the terms occupying the odd places, find the common ratio}$
$\displaystyle \text{of the G.P.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle \text{Let the G.P. contain }2n\text{ terms.}$
$\displaystyle \text{The terms occupying the odd places are}$
$\displaystyle a,ar^2,ar^4,\ldots,ar^{2n-2}.$
$\displaystyle \text{Let their sum be }S.$
$\displaystyle \therefore S=a+ar^2+ar^4+\ldots+ar^{2n-2}.$
$\displaystyle \text{The terms occupying the even places are}$
$\displaystyle ar,ar^3,ar^5,\ldots,ar^{2n-1}.$
$\displaystyle \text{Their sum is}$
$\displaystyle ar+ar^3+ar^5+\ldots+ar^{2n-1}$
$\displaystyle =r(a+ar^2+ar^4+\ldots+ar^{2n-2})$
$\displaystyle =rS.$
$\displaystyle \therefore \text{Sum of all the terms}=S+rS=(1+r)S.$
$\displaystyle \text{Given, the sum of all the terms is }5\text{ times the sum of the odd-place terms.}$
$\displaystyle (1+r)S=5S$
$\displaystyle 1+r=5$
$\displaystyle \therefore r=4.$
$\displaystyle \therefore \text{The common ratio of the G.P. is }4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Let }a_n\text{ be the }n^{\mathrm{th}}\text{ term of a G.P. of positive numbers. Let}$
$\displaystyle \sum\limits_{n=1}^{100}a_{2n}=\alpha\qquad\text{and}\qquad\sum\limits_{n=1}^{100}a_{2n-1}=\beta,$
$\displaystyle \text{where }\alpha\neq\beta.\text{ Prove that the common ratio of the G.P. is }\frac{\alpha}{\beta}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle \text{For every positive integer }n,$
$\displaystyle a_{2n}=ar^{2n-1}$
$\displaystyle \text{and}\qquad a_{2n-1}=ar^{2n-2}.$
$\displaystyle \therefore a_{2n}=r\,a_{2n-1}.$
$\displaystyle \text{Hence,}$
$\displaystyle \alpha=\sum\limits_{n=1}^{100}a_{2n}$
$\displaystyle =\sum\limits_{n=1}^{100}r\,a_{2n-1}$
$\displaystyle =r\sum\limits_{n=1}^{100}a_{2n-1}$
$\displaystyle =r\beta.$
$\displaystyle \therefore r=\frac{\alpha}{\beta}.$
$\displaystyle \text{Since the terms are positive, }\beta>0,\text{ so the division is valid.}$
$\displaystyle \therefore \text{The common ratio of the G.P. is }\frac{\alpha}{\beta}.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Find the sum of }2n\text{ terms of a series in which every even term is }a$
$\displaystyle \text{times the preceding term and every odd term is }c\text{ times the preceding term,}$
$\displaystyle \text{the first term being unity.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the terms of the series be }a_1,a_2,a_3,\ldots,a_{2n}.$
$\displaystyle \text{Given, }a_1=1.$
$\displaystyle a_2=a\cdot a_1=a,$
$\displaystyle a_3=c\cdot a_2=ac,$
$\displaystyle a_4=a\cdot a_3=a^2c,$
$\displaystyle a_5=c\cdot a_4=a^2c^2,\ldots$
$\displaystyle \therefore S_{2n}=1+a+ac+a^2c+a^2c^2+\ldots\text{ to }2n\text{ terms}.$
$\displaystyle \text{Grouping the terms in pairs,}$
$\displaystyle S_{2n}=(1+a)+ac(1+a)+(ac)^2(1+a)+\ldots+(ac)^{n-1}(1+a)$
$\displaystyle =(1+a)\left[1+ac+(ac)^2+\ldots+(ac)^{n-1}\right].$
$\displaystyle \text{If }ac\neq1,$
$\displaystyle S_{2n}=(1+a)\left[\frac{1-(ac)^n}{1-ac}\right]$
$\displaystyle =(1+a)\left[\frac{(ac)^n-1}{ac-1}\right].$
$\displaystyle \text{If }ac=1,\text{ then}$
$\displaystyle S_{2n}=n(1+a).$
$\displaystyle \therefore \text{The required sum is }(1+a)\frac{1-(ac)^n}{1-ac},\quad ac\neq1.$
$\displaystyle \\$