Question

Find the first three terms of the infinite geometric sequence with sum $81$ and common ratio $\frac{1}{3}$.

Answer 1

Let the first term in Geometric infinite series be $a$
Given, common ratio is $\frac{1}{3}$ and sum is $81$
Sum is $81=a+\frac{a}{3}+.........=a(1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+.............)=a\left(\frac{1}{1-1/3}\right)=\frac{3a}{2}$
Therefore, $a=81\times \frac{2}{3}=54$
So, the first three terms are $54,18,6$.