Question

Let α and β be the roots of equation $x^2-7x-3=0.\text{If}{a}_n={\alpha }_n-{\beta }_{n}forn\ge 1,\text{then value of}\frac{a_{10}-3a_8}{3a_9}$ is

• A. $\frac{7}{3}$
• B. $\frac{-7}{3}$
• C. $\frac{3}{7}$
• D. $\frac{-3}{7}$

Answer 1

The correct option is A $\frac{7}{3}$

Given that α and β are roots of $x^2-7x-3=0.$

Thus, ${\alpha }^2-7\alpha -3=0\text{and}{\beta }^2-7\beta -3=0$

$\Rightarrow {\alpha }^2-3=7\alpha \text{and}{\beta }^2-3=7\beta$

Since, $a_n={\alpha }^n-{\beta }^n,n\ge 1$

$\Rightarrow a_{10}={\alpha }^{10}-{\beta }^{10},a_8={\alpha }^8-{\beta }^8\text{and}{a}_9={\alpha }^9-{\beta }^9.$

$=\frac{{\alpha }^8\times 7\alpha -{\beta }^8\times 7\beta }{3({\alpha }^9-{\beta }^9)}$ [from (i)]

$=\frac{7({\alpha }^9-{\beta }^9)}{3({\alpha }^9-{\beta }^9)}=\frac{7}{3}$

Consider, $\frac{a_{10}-3a_8}{3a_9}$

$=\frac{({\alpha }^{10}-{\beta }^{10})-3({\alpha }^8-{\beta }^8)}{3({\alpha }^9-{\beta }^9)}$

$=\frac{{\alpha }^8({\alpha }^2-3)-{\beta }^8({\beta }^2-3)}{3({\alpha }^9-{\beta }^9)}$

Hence, the correct answer is option (a).