Sequences And Series Question 11
Question 11 - 2024 (29 Jan Shift 2)
If each term of a geometric progression $a_{1}, a_{2}, a_{3}, \ldots$ with $\mathrm{a}{1}=\frac{1}{8}$ and $\mathrm{a}{2} \neq \mathrm{a}{1}$, 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
(1) $2^{15}$
(2) $-2^{18}$
(3) $2^{18}$
(4) $-2^{15}$
Show Answer
Answer (4)
Solution
Let $r^{\prime}$ th term of the GP be $a r^{n-1}$. Given,
$2 a_{r}=a_{r+1}+a_{r+2}$
$2 a r^{n-1}=a r^{n}+a r^{n+1}$
$\frac{2}{r}=1+r$
$r^{2}+r-2=0$
Hence, we get, $r=-2($ as $r \neq 1)$
So, $\mathrm{S}{20}-\mathrm{S}{18}=$ (Sum upto 20 terms) - (Sum upto
18 terms $)=\mathrm{T}{19}+\mathrm{T}{20}$
$\mathrm{T}{19}+\mathrm{T}{20}=\operatorname{ar}^{18}(1+\mathrm{r})$
Putting the values $\mathrm{a}=\frac{1}{8}$ and $\mathrm{r}=-2$;
we get $T_{19}+T_{20}=-2^{15}$
