JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 16)
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
$$-2^{15}$$
$$2^{15}$$
$$-2^{18}$$
$$2^{18}$$
Explanation
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 $$T_{19}+T_{20}=-2^{15}$$
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