Question

What is the sum of the first five terms of the geometric sequence in which a1=10 and r=1/2?

Answer 1

(10(1-(1/2)^5)/1-(1/2) = 

20(1-1/32)
=155/8

Answer 2

Answer:

The sum of the first five terms of the geometric sequence whose first term is 10 and common ratio is  $\frac{1}{2}$ is $19\frac{3}{8}$

Step-by-step explanation:

A geometric sequence is a sequence  where each term is find  by multiplying the previous term by a constant non-zero number known as the common ratio.

Here, given $a_1=10$ and $r=\frac{1}{2}$

We have to find the sum of the first five terms of the geometric sequence whose first term is 10 and common ratio is  $\frac{1}{2}$.

Sum of a geometric sequence is given as :

$S_n=\frac{a_1(1-r^n)}{1-r}$

Substitute the values,

$S_n=\frac{10(1-(\frac{1}{2})^5)}{1-\frac{1}{2}}$

Solving , we get

$\Rightarrow S_n=\frac{10(1-(\frac{1}{2})^5)}{\frac{1}{2}}$

$\Rightarrow S_n=\frac{10(1-\frac{1}{32})}{\frac{1}{2}}$      

$\Rightarrow S_n=\frac{10(\frac{32-1}{32})}{\frac{1}{2}}$    

$\Rightarrow S_n=\frac{10(\frac{32-1}{32})}{\frac{1}{2}}$  

$\Rightarrow S_n=20(\frac{31}{32})$  

$\Rightarrow S_n=10(\frac{31}{16})$  

$\Rightarrow S_n=\frac{310}{16}=\frac{155}{8}$  

$\Rightarrow S_n=19\frac{3}{8}$  

Thus, the sum of the first five terms of the geometric sequence whose first term is 10 and common ratio is  $\frac{1}{2}$ is $19\frac{3}{8}$