Question

Find the $9^{th}$ term and the general term of the progression $\frac{1}{4},-\frac{1}{2},1,-2,...$

• A. $64,{\left(-1\right)}^{n-1}{2}^{n-3}$
• B. $63,{\left(-1\right)}^{n-1}{2}^{n-2}$
• C. $62,{\left(-1\right)}^{n-2}{2}^{n-3}$
• D. $61,{\left(-2\right)}^{n-1}{2}^{n-3}$

Answer 1

Answer: (A)

The given progression is clearly a $G.P$. with first term $a=1/4$ and common ratio $r=-2$.
$\therefore 9^{th}$ term $=a_9$
$=a{r}^{\left(9-1\right)}$
$=\frac{1}{4}{\left(-2\right)}^8$
$=64$
and, General term $=a_n$
$=a{r}^{\left(n-1\right)}$
$=\frac{1}{4}{\left(-2\right)}^{n-1}$
$={\left(-1\right)}^{n-1}{2}^{n-3}$