Solution:

Let the first A.P have first term a and common difference d.
The sum of n terms of this A.P is given by Sn = n/2 [2a + (n-1)d]

Let the second A.P have first term A and common difference D.
The sum of n terms of this A.P is given by SN = n/2 [2A + (n-1)D]

Given, (Sn/SN) = (7n-5)/(5n+17)

=> [(n/2) (2a + (n-1)d) ] / [(n/2) (2A + (n-1)D)] = (7n-5)/(5n+17)

=> [2a + (n-1)d] / [2A + (n-1)D] = (7n-5)/(5n+17)

=> 10an - 5a + 17dn - 7d = 14An - 5A + 17DN - 7D

=> 10an + 17dn + 5A + 7D = 14An + 17DN + 5a + 7d

=> 10(a + 3d) + 5A + 7D = 14(A + 3D) + 5a + 7d

=> 5a + 7d = 5A + 7D

=> 5(a - A) = 7(D - d)

Since a - A and D - d are constants, we can conclude that the ratio of corresponding terms of the two A.Ps is constant. Hence, the ____ term of the two series are equal.

Therefore, the answer is (d) None.

Explanation:
- Introduction to the problem and given information
- Derivation of the formula for sum of n terms of two A.Ps
- Calculation of the ratio of the sums of n terms of the two A.Ps
- Simplification of the ratio to obtain a relation between the corresponding terms of the two A.Ps
- Conclusion that the corresponding terms have a constant ratio and hence the answer is None.