JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q26

MathematicsInteger+4 / -1Question 26 of 90

For a natural number n, let $\alpha_n=19^n-12^n$ . Then, the value of $\dfrac{31\alpha_9-\alpha_{10}}{57\alpha_8}$ is ______.

Your answer

Answer and worked solution

Official answer: 4

Recurrence. 19 and 12 are roots of $t^2-31t+228=0$ , so $\alpha_{10}=31\alpha_9-228\alpha_8$ .

Value. $\dfrac{31\alpha_9-\alpha_{10}}{57\alpha_8}=\dfrac{228}{57}=4$ .

Answer: 4

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 26