Let $P_{n}=\alpha^{n}+\beta^{n}, n \in \mathbf{N}$. If $P_{10}=123, P_{9}=76$, $\mathrm{P}_{8}=47$ and $\mathrm{P}_{1}=1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :

(A) $\mathrm{x}^{2}-\mathrm{x}+1=0$

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Sequences and Series

Let

P_{n} = α^{n} + β^{n}, n \in N

. If

P_{10} = 123, P_{9} = 76

P_{8} = 47

and

P_{1} = 1

, then the quadratic equation having roots

\frac{1}{α}

and

\frac{1}{β}

is :

Answer: (B)

Observe that

P_{10} = P_{9} + P_{8}

(123 = 76 + 47), suggesting a recurrence relation

P_{n} = P_{n - 1} + P_{n - 2}

.

This means

α^{n} + β^{n} = α^{n - 1} + β^{n - 1} + α^{n - 2} + β^{n - 2}

.

Dividing by

α^{n - 2} β^{n - 2}

and simplifying, we get that

α

and

β

satisfy

x^{2} = x + 1

, or

x^{2} - x - 1 = 0

.

From Vieta's formulas:

α + β = 1

and

α β = - 1

.

For the equation with roots

\frac{1}{α}

and

\frac{1}{β}

Sum of roots: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{1}{- 1} = - 1$
Product of roots: $\frac{1}{α β} = \frac{1}{- 1} = - 1$

The equation is

x^{2} - (- 1) x + (- 1) = 0

, i.e.,

x^{2} + x - 1 = 0

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option