Question

Let $\alpha_{1}, \beta_{1}$ be the roots of $x^{2}-6 x+p=0$ and $\alpha_{2}, \beta_{2}$ be the roots of $x^{2}-54 x+q=0$. 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

Let $\alpha_{1}= A , \beta_{1}= AR , \alpha_{2}= AR ^{2}, \beta_{2}= AR ^{3}$
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 $\alpha_2 + \beta_2 = 54 \Rightarrow AR^2 ( 1 + R) = 54...$ (3)
$\alpha_2\beta_2 = q \Rightarrow A^2R^2 = q ... $(4)
Now, on dividing Eq. (3) by Eq. (1), we get
$\frac{ AR ^{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, $q - p =\frac{2187-27}{4}=\frac{2160}{4}=540$