JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 5

α. The terms pair up as $(5k-2)+(5k-1)=10k-3$ for $k=1,\dots,20$:
$$\alpha=10\cdot210-60=2040,\qquad\frac\alpha{1020}=2.$$

β. $(\tan\beta)^2$ is a root of $x^2+x-2=0$ (roots 1 and −2), so $\tan^2\beta=1$ and $\beta=\tfrac\pi4$.

Value: $\tfrac12+\tfrac32=2$.

(A)