**Solution:**

Given that,

S_{7} = 49

S_{17} = 289

We know the sum of n terms;

S_{n} = n /2 [2a + (n – 1) d ]

Therefore,

S_{7} = 7 /2 [2a + (n – 1) d ]

S_{7} = 7/2 [2a + (7 – 1) d ]

49 = 7/2 [2a + 6 d ]

7 = (a + 3d)

a + 3d = 7 …………………………………. **(i)**

Similarly,

S_{17} = 17/2 [2a + (17 – 1) d ]

289 = 17/2 (2a + 16 d )

17 = ( a + 8d )

a + 8d = 17 ………………………………. **(ii)**

Subtracting equation **(i)** from equation **(ii)**,

5 d = 10

d = 2

From equation **(i)**,

a + 3(2) = 7

a + 6 = 7

a = 1

Hence,

S_{n} = n /2[2a + (n – 1) d ]

= n /2[2(1) + (n – 1) × 2]

= n /2(2 + 2n – 2)

= n /2(2n)

= n^{2}