Question

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

Answer 1

Given:

`(4)/(3),(2)/(3),(1)/(3),(1)/(6)`

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

`\begin{gathered} r=\frac{\text{second term}}{\text{first term}}=(2)/(3)/(4)/(3) \\ r=(2)/(3)*(3)/(4) \\ r=(1)/(2) \end{gathered}`

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

`\begin{gathered} S_(\infty)=(a)/(1-r)\textr \\ \\ \text{where, a is the first term} \\ a=(4)/(3) \\ \\ \text{Hence,} \\ S_(\infty)=(a)/(1-r) \\ S_(\infty)=((4)/(3))/(1-(1)/(2)) \\ S_(\infty)=((4)/(3))/((1)/(2)) \\ S_(\infty)=(4)/(3)/(1)/(2) \\ S_(\infty)=(4)/(3)*(2)/(1) \\ S_(\infty)=(8)/(3) \\ S_(\infty)=2(2)/(3) \end{gathered}`

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

`(8)/(3)\text{ OR 2}(2)/(3)`