JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q69

MathematicsSingle correct+4 / -1Question 69 of 90

If the sum of the first n terms of the series $\sqrt3+\sqrt{75}+\sqrt{243}+\sqrt{507}+\dots$ is $435\sqrt3$ , then n equals :

Answer and worked solution

Official answer: B

The terms are $\sqrt3(1+5+9+13+\dots)$ , an AP with difference 4, whose sum is $\sqrt3\,n(2n-1)$ . So $2n^2-n-435=0$ and $n=15$ .

JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 69