Number Theory problem 3 by Dhaval Furia

$(2n+1)+(2n+3)+(2n+5)+\cdots+(2n+47)=5280$

$24(2n)+(1+3+5+\cdots+45+47)=5280$

Since the sum of first $n$ odd numbers is $n^2$ ,

$24(2n)+(24^2)=5280$

$24(2n)+576=5280$

$48n)=4704$

$n=98$

Since the sum of first $n$ natural numbers is $\dfrac{n(n+1)}{2}$ ,

$\implies1+2+3+4\cdots+97+98= \dfrac{98(99)}{2}$

$\implies1+2+3+4\cdots+97+98= \boxed{4851}$

$\begin{aligned} \underbrace{\overbrace{(2n+1)}^a+(2n+3)+(2n+5)+\cdots + \overbrace{(2n+47)}^l}_{n=24 \text{ terms}} & = 5280 & \small \blue{\text{Sum of AP, }S = \frac {n(a+l)}2} \\ \frac {24(2n+1+2n+47)}2 & = 5280 \\ 12 \times 4 (n+12) & = 5280 \\ \implies n & = \frac {5280}{48} - 12 = 98 \\ \implies 1 + 2 + 3 + \cdots + n & = \blue{\frac {n(n+1)}2} & \small \blue{\text{Sum of AP}} \\ & = \frac {98(99)}2 = \boxed{4851} \end{aligned}$