Question

Match elements of List I with elements in List II.

Answer 1

a] Let the roots be $a-d,a,a+d$
$3a=9\Rightarrow a=3$
$a^2-ad+a^2+ad+a^2-d^2=26$
$3a^2-d^2=26$
$3\left(9\right)-d^2=26$
$\Rightarrow d=1$
$\therefore 2,3,4$
Product $=k=24$
b] Let the roots be $\frac{a}{r},a,ar$
$a^3=64\Rightarrow a=4$
$\frac{a}{r}+a+ar=14\Rightarrow \frac{1}{r}+r=\frac{5}{2}$
$\Rightarrow r=2$ or $r=\frac{1}{2}$
$k=\frac{a^2}{r}+a^{2}r+a^2=8+32+16=56$

c] Let the roots be $\frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}$
$a-d+a+a+d=\frac{\frac{6}{6}}{\frac{1}{6}}=6$
$a=2$
Hence, $\frac{1}{2}$ is a root of the equation.
$\Rightarrow \frac{6}{8}-\frac{k}{4}+3-1=0$
$\Rightarrow \frac{11}{4}=\frac{k}{4}$
$k=11$
$3k=33$
d) Let the roots be $\alpha ,\beta ,\gamma$
$\alpha +\beta +\gamma =0$
It is given that $\alpha +\beta =3$
Hence, $\gamma =-3$