Question

Let $a^2,a(b-3),4b$ be the first three terms of an increasing A.P. where $a,b\in I$ and $5\le b\le 15$.
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)

$\Theta a^2,a(b-3),4b$ are in A.P.
$a^2+4b=2a(b-3)$
$\Rightarrow a^2-2a(b-3)+4b=0$
$a=\frac{2(b-3)\pm \sqrt{4(b-3{)}^2-4\cdot 4b}}{2}=b-3\pm \sqrt{b^2-10b+9}=b-3\pm \sqrt{(b-1)(b-9)}$
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 $16,28,40,\dots \dots \dots (a=4)$
$100,70,40,\dots \dots (a=10)($ rejected $)$
$\frac{b}{a}=\frac{10}{4}=\frac{5}{2}$