Home -> Solved problems -> Find the general term of the sequence

### Solution

To start, we know that we are dealing with a geometric sequence, thus
$a_{2}=r a_{1}=-\frac{2}{3}\;\;\;\;\;\;\;(1)$
$$r$$ is the common ratio.
An infinite geometric series converges if and only if $$|r|<1$$, therefore
\begin{aligned} \lim _{n \rightarrow\infty} \sum_{k=1}^{n} a_{k}&=\frac{a_{1}}{1-r} \\\\ \Rightarrow \frac{a_{1}}{1-r}&=\frac{3}{5}\;\;\;\;\;\;\;(2) \end{aligned}
$$(1)$$ and $$(2)$$, gives
\begin{aligned} &\frac{a_{1}}{1+\frac{2}{3 a_{1}}}=\frac{3}{5} \\\\ &\frac{a_{1}}{\frac{3 a_{1}+2}{3 a_{1}}}=\frac{3}{5} \\\\ &\frac{3 a_{1}^{2}}{3 a_{1}+2}=\frac{3}{5} \\\\ &3 a_{1}^{2}=\frac{9}{5} a_{1}+\frac{6}{5} \\\\ &3 a_{1}^{2}-\frac{9}{5} a_{1}-\frac{6}{5}=0 \\\\ &15 a_{1}^{2}-9 a_{1}-6=0 \\\\ &15\left(a_{1}-1\right)\left(a_{1}+\frac{6}{15}\right)=0 \\\\ &a_{1}=1\;\;\;\; or\;\;\;\; a_{1}=-\frac{6}{15} \end{aligned}
Let’s discuss the found values
\begin{aligned} &a_{1}=1\;\;\;\; gives \;\;\;\;r=-\frac{2}{3}\\\\ &a_{1}=-\frac{6}{15}\;\;\;\; gives \;\;\;\;r=\frac{5}{3}>1\;\;\;\;\text { not possible because }|r|<1\\\\ &\text { So we consider the results } a_{1}=1 \text { and } r=-\frac{2}{3} \end{aligned}
The $$n$$-th term of a geometric sequence with first term $$a_{1}$$ and common ratio $$r$$ is given by
\begin{aligned} &a_{n}=a_{1} r^{n-1} \\\\ &a_{n}=\left(-\frac{2}{3}\right)^{n-1} \quad n \geqslant 1 \end{aligned}

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Home -> Solved problems -> Find the general term of the sequence

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