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 $a _1=\frac{1}{8}$ 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
(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, $S _{20}-S _{18}=$ (Sum upto 20 terms) - (Sum upto
18 terms $)=T _{19}+T _{20}$
$T _{19}+T _{20}=\operatorname{ar}^{18}(1+r)$
Putting the values $a=\frac{1}{8}$ and $r=-2$;
we get $T _{19}+T _{20}=-2^{15}$
