If the sum of the first $7$ terms of an AP is $49$ and that of the first $17$ terms is $289$, what is the sum of the first $n$ terms?

For an AP, $S_n=\frac{n}{2}[2a+(n-1)d]$. Given $$\frac{7}{2}(2a+6d)=49 \Rightarrow a+3d=7$$ and $$\frac{17}{2}(2a+16d)=289 \Rightarrow a+8d=17.$$ Subtracting gives $5d=10$, so $d=2$. Then $a+6=7$, hence $a=1$. Therefore $$S_n=\frac{n}{2}[2+(n-1)2]=\frac{n}{2}(2n)=n^2.$$