Question

Sol. [Key Concept: The series (two interleaved APs) sums to $\alpha = 2040$, so $(tan\beta)^{\alpha/1020} = \tan^{2}\beta$ equals the root $1$ of $x^{2} + x - 2 = 0$; hence $\sin^{2}\beta + 3\cos^{2}\beta = 2$.]

$a = (3 + 8 + 13\ldots$ upto 20 terms $) + (4 + 9 + 14 +$ ... upto 20 terms)

$${= \frac{20}{2}\lbrack 6 + 19 \times 5\rbrack + \frac{20}{2}(8 + 19 \times 5) }{= 2040 }{(tan\beta)^{\frac{2040}{1020}} = \tan^{2}\beta }{x^{2} + x - 2 = 0 < \ _{- 2}^{1} }{\Rightarrow \tan^{2}\beta = 1;\sin^{2}\beta = \frac{1}{2}\&\cos^{2}\beta = \frac{1}{2} }{\sin^{2}\beta + 3\cos^{2}\beta = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2 }$$