Question

The first four terms of a sequence are given by $T _{1}=0, T _{2}=1, T _{3}=1, T _{4}=2$. The general term is given by $T_{ n }=A \alpha^{ n -1}+B \beta^{ n -1}$ where $A, B, \alpha, \beta$ are independent of $n$ and $A$ is positive.
The quadratic equation whose roots are $\alpha$ and $\beta$ is given by

• A. $x^{2}-2 x-1=0$
• B. $x^{2}-2 x-2=0$
• C. $x^{2}-x-1=0$
• D. None

Answer 1

Answer: (C)

We have
$ T _{1}=0\,\,\, ...(i)$
$\Rightarrow A+B=0$
$T_{2}=1\,\,\, ...(ii)$
$\Rightarrow A \alpha+ B \beta=1$
$T _{3}= A \alpha^{2}+ B \beta^{2}\,\,\, ...(iii)$
$=1$
$T_{4}=A \alpha^{3}+B \beta^{3}=2 \,\,\, ...(iv)$
From (i), we get $A = - B$

$\therefore \alpha$ and $\beta$ are roots of the equation $x^{2}-x-1=0$
$\Rightarrow \alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$
As $(\alpha-\beta)=\sqrt{5}$
so from $A (\alpha-\beta)=1$,
we get $A=\frac{1}{\sqrt{5}}$ and $B=\frac{-1}{\sqrt{5}}($ As $B=-A)$
Hence $5\left( A ^{2}+ B ^{2}\right)=2$