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

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, $\frac{\frac{ a + ar ^2}{2}}{ ar }=\frac{5}{3}$
$\Rightarrow 3+3 r^2=10 r \Rightarrow(r-3)(3 r-1)=0 \Rightarrow r=3 \text { or } \frac{1}{3} $.....(4)
$\therefore \text { On solving (3) and (4), we get }$
$ b(r=3)=3 $