If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Given S7=49 and S17=289
By using Sn=n/2[2a+(n−1)d] we have,
S7=7/2[2a+(7−1)d]=49
⟹49=7/2[2a+(7−1)d]
⟹49=7/2(2a+6d)
⟹7=a+3d
⟹a+3d=7.. (i)
S17=17/2[2a+(17−1)d]=289
⟹289=17/2[2a+(17−1)d]
⟹289=17/2(2a+16d)
⟹17=a+8d
⟹a+8d=17. (ii)
Substituting (i) from (ii), we get
5d=10 or d=2
From equation (i),
a+3(2)=7
a+6=7 or a=1
Sn=n/2[2(1)+(n−1)2]=n/2[2+(n−1)2]=n/2(2+2n−2)=n²