If each term of a geometric progression a_1,a_2,a_3, with a_1=18 and a_2 a_1 is the arithm…

Question

If each term of a geometric progression $a_1,a_2,a_3,\ldots$ with $a_1=\dfrac18$ and $a_2
e a_1$ 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:

Accepted Answer

The condition $a_n = \dfrac{a_{n+1}+a_{n+2}}{2}$ gives $1 = \dfrac{r+r^2}{2} \Rightarrow r^2+r-2=0 \Rightarrow (r-1)(r+2)=0$. Since $a_2\ne a_1$, $r=-2$.

$$S_{20}-S_{18} = a_{19}+a_{20} = a_1r^{18}(1+r) = \frac18(-2)^{18}(-1) = -\frac{2^{18}}{8} = -2^{15}.$$

Solutions · 1