Mathematics - Sequence & Series Question with Solution | TestHub

Put $\mathrm{b}=\mathrm{ar},\mathrm{c}=a{r}^2$. The mean conditions give $2\mathrm{ab}/(\mathrm{a}+\mathrm{b})=12$ and $(\mathrm{b}+\mathrm{c})/2=3$. Solving with $\mathrm{r}\ne 0$ gives $\mathrm{r}=-2$ and $(\mathrm{a},\mathrm{b},\mathrm{c})=(3,-6,12)$. Their mean is 3, b=-6, and no term $3(-2{)}^{\wedge }(\mathrm{n}-1)$ is an integer square.