Solution:

Provided here that,

S₇ = 49

S₁₇ = 289

We are aware that, Sum of n terms;

S_n = n/2 [2a + (n – 1)d]

As a result,

S₇= 7/2 [2a +(n -1)d]

S₇ = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a + 6d]

7 = (a+3d)

a + 3d = 7 …………………………………. (i)

Similarly,

S₁₇ = 17/2 [2a+(17-1)d]

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ………………………………. (ii)

Subtracting eq. (i) from eq. (ii),

5d = 10

d = 2

From eq. (i), we can rewrite it as;

a+3(2) = 7

a+ 6 = 7

a = 1

As a result,

S_n = n/2[2a+(n-1)d]

= n/2[2(1)+(n – 1)×2]

= n/2(2+2n-2)

= n/2(2n)

= n²