Question

The first three terms of an arithmetic-geometric progression are $3,-1$ and $-1.$ The next term of the progression is

• A. $2$
• B. $-2$
• C. $\frac{-5}{27}$
• D. $\frac{-5}{9}$

Answer 1

Answer: (D)

Let the first four terms of the $A.G.P.$ be
$3,\left(3+d\right)r,\left(3+2d\right)r^2,\left(3+3d\right)r^3$
Hence, $\left(3+d\right)r=-1$ and $\left(3+2d\right)r^2=-1$
$\frac{{\left(3+d\right)}^2{r}^2}{\left(3+2d\right)r^2}=\frac{1}{-1}$
$\Rightarrow 9+d^2+6d=-3-2d$
$\Rightarrow d^2+8d+12=0\Rightarrow d=-2,-6$
$r=\frac{-1}{3+d}\Rightarrow r=-1$ or $\frac{1}{3}$
(i) for $d=-2,r=-1;$ next term is
$\left(3+3d\right)r^3=\left(3-6\right){\left(-1\right)}^3=3$
(ii) for $d=-6,r=\frac{1}{3};$ next term is
$\left(3+3d\right)r^3=\left(3-18\right){\left(\frac{1}{3}\right)}^3=\frac{-15}{27}=\frac{-5}{9}$
So, the next term is $3$ or $\frac{-5}{9}$