Question

Let $a^2, a(b-3), 4 b$ be the first three terms of an increasing A.P. where $a, b \in I$ and $5 \leq b \leq 15$.
The value of $\left(\frac{b}{a}\right)$ is equal to

• A. 1
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. 7

Answer 1

Answer: (C)

$\Theta a^2, a(b-3), 4 b$ are in A.P.
$a^2+4 b=2 a(b-3) $
$\Rightarrow a^2-2 a(b-3)+4 b=0$
$a=\frac{2(b-3) \pm \sqrt{4(b-3)^2-4 \cdot 4 b}}{2}=b-3 \pm \sqrt{b^2-10 b+9}=b-3 \pm \sqrt{(b-1)(b-9)}$
Since, $a$ is an integer and $b \in[5,15]$
$\therefore$ Possible values of $b$ are $9 \& 10$.
for $b=9, a=6$ and A.P. is $36,36,36$ (rejected)
for $b=10, a=10,4$ A.P. is $16,28,40, \ldots \ldots \ldots(a=4)$
$100,70,40, \ldots \ldots( a =10)($ rejected $)$
$\frac{ b }{ a }=\frac{10}{4}=\frac{5}{2}$