The sum of first 7 terms of an AP is 49 and the sum of its first 17 terms is 289.

Let a be the first term and d be the common difference of the given AP.
Then we have:

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{7}=\frac{7}{2}[2 a+6 d]=7[a+3 d]$

$S_{17}=\frac{17}{2}[2 a+16 d]=17[a+8 d]$

However, S7 = 49 and S17 = 289
Now, 7[a + 3d] = 49 
⇒ a + 3d = 7           ...(i)
Also, 17[a + 8d] = 289  
​⇒ a + 8d = 17           ...(ii)

Subtracting (i) from (ii), we get:

5d = 10
⇒ d = 2

Putting d = 2 in (i), we get:
a + 6 = 7 
⇒ a = 1
Thus, a = 1 and d = 2

$\therefore$ Sum of $n$ terms of AP $=\frac{n}{2}[2 \times 1+(n-1) \times 2]=n[1+(n-1)]=n^{2}$