Mathematics - Sequence & Series Question with Solution | TestHub

$a+a{r}^{n-1}=66$ and $ar\cdot a{r}^{n-2}=a^2{r}^{n-1}=128$, so $a\cdot a{r}^{n-1}=128$. So $a$ and $a{r}^{n-1}$ are roots of $t^2-66t+128=0$, i.e. 2 and 64. Increasing: $a=2,a{r}^{n-1}=64$. $\frac{a\left(r^n-1\right)}{r-1}=126$ gives $\frac{a{r}^n-a}{r-1}=\frac{64r-2}{r-1}=126$, so $r=2$, and $2\cdot 2^{n-1}=64$ gives $n=6$.