If each term of a geometric progression $a_1, a_2, a_3, \ldots$ with $a_1=\frac{1}{8}$ and $a_2 \neq a_1$, is the arithmetic mean of the next two terms and $S_n=a_1+a_2+\ldots . .+a_n$, then $S_{20}-S_{18}$ is equal to
Solution
<p>Let $r^{\prime}$th term of the GP be $a^{n-1}$. Given,</p>
<p>$$\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \\
& 2 a r^{n-1}=a r^n+a r^{n+1} \\
& \frac{2}{r}=1+r \\
& r^2+r-2=0
\end{aligned}$$</p>
<p>Hence, we get, $r=-2$ (as $r \neq 1$)</p>
<p>So, $\mathrm{S}_{20}-\mathrm{S}_{18}=$ (Sum upto 20 terms) $-$ (Sum upto 18 terms) $=\mathrm{T}_{19}+\mathrm{T}_{20}$</p>
<p>$\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})$</p>
<p>Putting the values $\mathrm{a}=\frac{1}{8}$ and $\mathrm{r}=-2$;</p>
<p>we get $T_{19}+T_{20}=-2^{15}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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