Question

Let ${\alpha }_1,{\beta }_1$ be the roots of $x^2-6x+p=0$ and ${\alpha }_2,{\beta }_2$ be the roots of $x^2-54x+q=0$. If ${\alpha }_1,{\beta }_1,{\alpha }_2,{\beta }_2$ form an increasing G.P., then sum of the digit of the value of $(q-p)$ is ______

Answer 1

Answer: 9

Let ${\alpha }_1=A,{\beta }_1=AR,{\alpha }_2=A{R}^2,{\beta }_2=A{R}^3$
we have ${\alpha }_1+{\beta }_1=6$
$\Rightarrow A(1+R)=6$ ... (1)
${\alpha }_1{\beta }_1=p\Rightarrow A^{2}R=p...$ (2)
Also ${\alpha }_2+{\beta }_2=54\Rightarrow A{R}^2(1+R)=\mathrm{54...}$ (3)
${\alpha }_2{\beta }_2=q\Rightarrow A^2{R}^2=q...$(4)
Now, on dividing Eq. (3) by Eq. (1), we get
$\frac{A{R}^2(1+R)}{A(1+R)}=\frac{54}{6}=9$
or $R^2=9$
$\therefore R=3$ (as it is an increasing G.P.)
$\therefore$ On putting $R=3$ in Eq. (1), we get
$A=\frac{6}{4}=\frac{3}{2}$
$\therefore p=A^{2}R=\frac{9}{4}\times 3=\frac{27}{4}$
and $q=A^2{R}^5=\frac{9}{4}\times 343=\frac{2187}{4}$
Hence, $q-p=\frac{2187-27}{4}=\frac{2160}{4}=540$