Question

If $(2n+1) + (2n+3) + (2n+5) + \dots + (2n+47) = 5280$, then what is the value of $1+2+3+\dots+n$? [TITA]

Accepted Answer

4851. The terms are of the form $2n + (2k-1)$ for $k = 1, 2, \dots, 24$ (since $2k - 1 = 47 \implies k = 24$). Sum $= 24 	imes 2n + (1 + 3 + 5 + \dots + 47) = 48n + 24^2 = 48n + 576$. $$48n + 576 = 5280$$ $$48n = 4704 \implies n = 98$$ We need $1 + 2 + \dots + 98 = \frac{98 	imes 99}{2} = 49 	imes 99 = 4851$.

Sum of AP and sum of first n natural numbers

Explanation

Practise this under exam conditions