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?
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Note: $\displaystyle\sum_{k=0}^{n-1}ar^k=a\dfrac{1-r^n}{1-r}$. Here $n=100$. – Paras Khosla Feb 20 '19 at 07:03
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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.
Ross Millikan
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You get $a_1\bigg(\frac{1-\color{blue}{(\frac23)^{100}}}{1-\frac23}\bigg)=15\implies a_1=\frac5{1-\color{blue}{(\frac23)^{100}}}$.