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Maths Question 17 – JEE-MAIN 2025

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Let Pn=αn+βn, n∈N. If P10=123, P9=76, P8=47 and P1=1, then the quadratic equation having roots 1α and 1β is:

For a sequence Pn=αn+βn, where α and β are roots of a quadratic equation x2−Ax+B=0, the sequence satisfies a linear recurrence relation.

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Ninja Strategy•Reciprocal Roots Property

Recognize that if ax2+bx+c=0 has roots α,β, then cx2+bx+a=0 has roots 1/α,1/β. First find the equation for α,β and then apply this transformation.

● Video Walkthrough
Step 1: Determine the sum and product of α and β✦ Active

The sequence Pn=αn+βn satisfies the recurrence relation Pn=(α+β)Pn−1−(αβ)Pn−2, where α and β are roots of x2−(α+β)x+αβ=0.

Given P1=α+β=1.

Using the recurrence relation for n=10: P10=(α+β)P9−(αβ)P8. Substituting P1=1:

P10=P9−(αβ)P8

Substitute the given values P10=123, P9=76, P8=47:

123=76−(αβ)(47)47=−47(αβ)αβ=−1

Thus, we have α+β=1 and αβ=−1.

Step 2: Form the quadratic equation with roots 1α and 1β○ Expand

Let the new roots be r1=1α and r2=1β.

The sum of the new roots is r1+r2=1α+1β=α+βαβ.

The product of the new roots is r1r2=1α⋅1β=1αβ.

Substitute the values of α+β=1 and αβ=−1:

Sum=1−1=−1 Product=1−1=−1

The quadratic equation with roots r1 and r2 is x2−(Sum of roots)x+(Product of roots)=0.

x2−(−1)x+(−1)=0x2+x−1=0
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