Question

The numbers a, $b$ and $c$ are between $2$ and $18$ , such that
(i) their sum is $25$
(ii) the numbers $2,a$ and $b$ are consecutive terms of an $A.P.$
(iii) the numbers $b,c,18$ are consecutive terms of a $G.P.$
The value of $abc$ is

• A. 500
• B. 450
• C. 720
• D. None of these

Answer 1

Answer: (D)

We have,
$a+b+c=25...(i)$
$2a=b+2...(ii)$
$c^2=18b...(iii)$
Eliminating a from (i) and (ii), we have
$b=16-\frac{2c}{3}$
Then from(iii),
$c^2=18\left(16-\frac{2c}{3}\right)$
or $c^2+12c-18\times 16=0$
or $(c-12)(c+24)=0$
Now, $c=-24$ is not possible since it does not lie between $2$ and $18$ .
Hence, $c=12$. Then from (iii), $b=8$ and finally from (ii), $a=5$
Thus, $a=5,b=8$ and $c=12$.
Hence, $abc=5\times 8\times 12=480$.
Also, equation $a{x}^2+bx+c=0$ is $5x^2+8x+12=0$,
which has imaginary roots.
If $a,b,c$ are roots of the equation $x^3+q{x}^2+rx+s=0$,
then the sum of product of roots taken two at a time is
$r=5\times 8+5\times 12+8\times 12=196.$