Question

For the geometric sequence 4/3,2/3,1/3,1/6,… find the infinite sum of the sequence

Answer 1

Solution:

Given:

$\frac{4}{3},\frac{2}{3},\frac{1}{3},\frac{1}{6}$

From the geometric sequence, the common ratio (r) is;

$\begin{gathered} r=\frac{\text{second term}}{\text{first term}}=\frac{2}{3}\div\frac{4}{3} \\ r=\frac{2}{3}\times\frac{3}{4} \\ r=\frac{1}{2} \end{gathered}$

To get the sum to infinity, the formula below is used;

$\begin{gathered} S_{\infty}=\frac{a}{1-r}\text{ when |r| < 1} \\ \\ \text{where, a is the first term} \\ a=\frac{4}{3} \\ \\ \text{Hence,} \\ S_{\infty}=\frac{a}{1-r} \\ S_{\infty}=\frac{\frac{4}{3}}{1-\frac{1}{2}} \\ S_{\infty}=\frac{\frac{4}{3}}{\frac{1}{2}} \\ S_{\infty}=\frac{4}{3}\div\frac{1}{2} \\ S_{\infty}=\frac{4}{3}\times\frac{2}{1} \\ S_{\infty}=\frac{8}{3} \\ S_{\infty}=2\frac{2}{3} \end{gathered}$

Therefore, the sum to infinity of the geometric sequence is;

$\frac{8}{3}\text{ OR 2}\frac{2}{3}$