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 $G.P.$
If $a , b$ and $c$ are roots of the equation $x ^{3}+ qx ^{2}+$ $rx + s =0$, then the value of $r$ is

• A. 184
• B. 196
• C. 224
• D. None of these

Answer 1

Answer: (B)

We have,
$a+b+c=25 \,\,\, ...(i)$
$2 a=b+2 \,\,\, ...(ii)$
$c^{2}=18 b\,\,\, ...(iii)$
Eliminating a from (i) and (ii), we have
$b =16-\frac{2 c }{3}$
Then from(iii),
$c ^{2}=18\left(16-\frac{2 c }{3}\right) $
or $c ^{2}+12 c -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, $a b c=5 \times 8 \times 12=480$.
Also, equation $a x^{2}+b x+c=0$ is $5 x^{2}+8 x+12=0$,
which has imaginary roots.
If $a , b , c$ are roots of the equation $x ^{3}+ qx ^{2}+ rx + s =0$,
then the sum of product of roots taken two at a time is
$r =5 \times 8+5 \times 12+8 \times 12=196 .$