If each term of a geometric progression [a_1, a_2, a_3, ] with [a_1=(1)/(8)] and [a_2 a_1]…

Question

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

Accepted Answer

Let $$r^{\prime}$$th term of the GP be $a^{n-1}$. Given,

$$\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \
& 2 a r^{n-1}=a r^n+a r^{n+1} \
& \frac{2}{r}=1+r \
& r^2+r-2=0
\end{aligned}$$

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 $$\mathrm{a}=\frac{1}{8}$$ and $$\mathrm{r}=-2$$;

we get $T_{19}+T_{20}=-2^{15}$

Solutions · 1