Question

IThe numbers $a,b$ and $c$ are between $2$ and $18$ such that
$\left(i\right)$ their sum is $25$
$\left(ii\right)$ the numbers $2,a,b$ are in A.P
$\left(iii\right)$ the numbers $b,c,18$ are consecutive terms of a G.P.

f $a,b,c$ are the roots of $x^3+q{x}^2+rx+s=0$ then the value of ${}^{\prime }{q}^{\prime }$ is

• A. $-25$
• B. $-20$
• C. $-13$
• D. $0$

Answer 1

Answer: (A)

$-25$

$\left(i\right)$ Sum$=a+b+c=25$ ........$\left(1\right)$
$\left(ii\right)$ The terms $2,a,b$ are in A.P
$\Rightarrow 2a=b+2$ .........$\left(2\right)$
$\left(iii\right)$ The terms $b,c,18$ are in G.P
$\Rightarrow c^2=18b$ ....$\left(3\right)$
From $\left(1\right)$ and $\left(2\right)$ $a+b+c=25$
$\Rightarrow a=25-b-c$ (on transposition from L.H.S to R.H.S)
$\Rightarrow \frac{b+2}{2}=25-b-c$ (from $\left(2\right)$)
$\Rightarrow b+2=50-2b-2c$ (Multiplying both sides by $2$)
$\Rightarrow b+2b=48-2c$
or $3b=48-2c$
From$\left(3\right)$ substituting for $b$ we get
$2c+3\frac{c^2}{18}=48$
$\Rightarrow 2c+\frac{c^2}{6}=48$
$\Rightarrow 12c+c^2-288=0$ (Multiplying by $6$)
$\Rightarrow c^2+12c-288=0$ (Rearranging)
$\Rightarrow (c-12)(c+24)=0$ (on factorizing the above equation)
Hence. $c=12,-24$$c=12$ is valid $(\because a,b,c)$ lie between $2$ and $18$
$\therefore c=12\Rightarrow c^2=18b$ (from $\left(3\right)$)
Substituting $c=12$ in $c^2=18b$
we get $b=\frac{144}{18}=8$
Substituting the values of $b=8,c=12$ in $a+b+c=25$
we get $a=25-20=5$
$\therefore a=5,b=8,c=12$
From $\left(1\right)$ we have $a+b+c=25=-q$
$\therefore -q=25$ or $q=-25$