Question

1.Find the sum of the first 20 terms of the geometric series

Answer 1

Final Answer: $\frac{15}{4} (1- \frac{1}{ 3^{20} } )$

Steps:
1) Given GP is :
$\frac{5}{2} + \frac{5}{6} + \frac{5}{18} +....20. terms$
First term, a = $\frac{5}{2}$
Common ratio, r = $\frac{1}{3}$
No. Of terms, n= 20

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

S(20) = 5/2(1-(1/3)^20)/(1-1/3)

= 5* 3/2*2 ( 1- (1/3)^20)

= 15/4 (1-(1/3)^20)

Answer 2

$\large{\mathbf{HELLO\:\: DEAR,}}$

$\mathbf{GIVEN \:\:THAT:-}$

$\mathbf{\frac{5}{2} + \frac{5}{6} + \frac{5}{18} +...... 20terms}$

where,

$\mathbf{a = \frac{5}{2} , r = \frac{\frac{5}{6}}{\frac{5}{2}} = \frac{1}{3}}$

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

$\mathbf{S_{20} = \frac{\frac{5}{2}(1 - (\frac{1}{3})^{20}}{1 - \frac{1}{3}}}$

$\mathbf{S_{20} = \frac{\frac{5}{2}(\frac{3^{20} - 1}{3^{20}})}{\frac{(3 - 1)}{3}}}$


$\mathbf{\frac{15(3^{20} - 1)}{4*3^{20}}}$


$\mathbf{15/4* (1 - 1/3^{20})}$


$\mathbf{S_{20} = 15/4* (1 - 1/3^{20})}$


$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$