Question

Find the sum of the following geometric progressions :

(i) 2, 6, 18, ..... to 7 terms

(ii) 1, 3, 9, 27, .... to 8 terms

(iii) $1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8},....$

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

(v) 4, 2, 1, $\frac{1}{2}....$ to 10 terms.

Answer 1

2, 6, 18, ..... to 7 terms

$a=2,r=\frac{6}{2}=3,n=7$

$S_n=a\frac{(r^n-1)}{r-1}$

$S_7=2\frac{(3^7-1)}{3-1}=\frac{2}{2}(3^7-1)$

$=2187-1=2186$

(ii) 1, 3, 9, 27, ..... to 8 terms

$(a=1,r=\frac{3}{1}=3,n=8)$

$S_n=a\frac{(r^n-1)}{r-1}$

$S_8=1\frac{(3^8-1)}{3-1}=3280$

(iii) $1,\frac{-1}{2},\frac{1}{4},\frac{-1}{8},.....9$ terms

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

$S_n=a\frac{(r^n-1)}{r-1}$

$S_9=1\frac{{\left(\frac{-1}{2}\right)}^9-1}{\frac{-1}{2}-1}$

$=\frac{\frac{-1}{512}-1}{\frac{-1}{2}-1}=\frac{\frac{-1-512}{512}}{\frac{-1-2}{2}}$

$=\frac{-513}{512}\times \frac{2}{-3}=\frac{171}{256}$

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

$a=a^2-b^2,r=\frac{a-b}{a^2-b^2}=\frac{1}{a+b},n=n$

$S_n=a\frac{(1-r^n)}{1-r}$ [$\therefore r<1$]

$S_n=(a^2-b^2)\frac{(1-\frac{1}{(a+b{)}^n})}{1-\frac{1}{a+b}}$

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

(v) $4,2,1,\frac{1}{2}.....10$ terms.

$a=4,r=\frac{2}{4}=\frac{1}{2},n=10$

$S_n=a\frac{(1-r^n)}{1-r}$

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

$=8(1-\frac{1}{2^{10}})=8(1-\frac{1}{1024})=\frac{1023}{128}$