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 $S_n=a_1+a_2+\cdots+a_n$, then $S_{20}-S_{18}$ is equal to
Select the correct option:
A
$2^{15}$
B
$-2^{18}$
C
$2^{18}$
D
$-2^{15}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Let $\mathrm{r}^{\prime}$ th term of the GP be ar ${ }^{\mathrm{n}-1}$. Given,
$$
\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \\
& 2 a^{n-1}=a r^n+a r^{n+1} \\
& \frac{2}{r}=1+r \\
& r^2+r-2=0
\end{aligned}
$$
Hence, we get, $r=-2($ as $r \neq 1)$
So, $\mathrm{S}_{20}-\mathrm{S}_{18}=$ (Sum upto 20 terms) - (Sum upto 18 terms ) $=\mathrm{T}_{19}+\mathrm{T}_{20}$
$$
\mathrm{T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})
$$
Putting the values $\mathrm{a}=\frac{1}{\mathrm{~s}}$ and $\mathrm{r}=-2$;
we get $T_{19}+T_{20}=-2^{15}$
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