Question

The first term of a geometric progression is equal to $b-2$, the third term is $b+6$, and the arithmetic mean of the first and third term to the second term is in the ratio $5:3$. Find the positive integral value of $b$.

Answer 1

Answer: 3

Let $I^{\text{st }}$ term $=a=b-2$....(1)
Third term $=a^2=b+6$, where $r$ is common ratio of G.P. ....(2)
$\therefore$ From (1) and (2), we get
$r^2=\frac{b+6}{b-2}$.....(3)
Also, $\frac{\frac{a+a{r}^2}{2}}{ar}=\frac{5}{3}$
$\Rightarrow 3+3r^2=10r\Rightarrow (r-3)(3r-1)=0\Rightarrow r=3\text{ or }\frac{1}{3}$.....(4)
$\therefore \text{ On solving (3) and (4), we get }$
$b(r=3)=3$