Solve geometric series equation with large terms

Question

Let $\{a_i\}$ is a geometric sequence with common ratio $r=2/3$. If $a_1+a_2+...+a_{100}=15$, $a_1+a_2+...+a_{99}$?

I think $a_1(1+\frac{2}{3}+...+(\frac{2}{3})^{99})=15 \implies a_1=5$, what wrong?

Answer 1

The sum $1+\frac{2}{3}+...+(\frac{2}{3})^{99} \neq 3$ The limit as the number of terms goes to infinity is $3$, but you have a finite sum here. I suspect your teacher is expecting an algebraic result, not a calculator one. It is very close, but not equal.

Answer 2

You get $a_1\bigg(\frac{1-\color{blue}{(\frac23)^{100}}}{1-\frac23}\bigg)=15\implies a_1=\frac5{1-\color{blue}{(\frac23)^{100}}}$.