Introduction:
In this problem, we are given two arithmetic progressions (A.P.s) and we need to find the term at which the sum of the first n terms is equal for both progressions. The ratio of the sums of the first n terms of the two A.P.s is given by (7n-5)/(5n+17).

Approach:
To solve this problem, we will use the formula for the sum of an arithmetic progression. The sum of the first n terms of an A.P. is given by the formula Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference of the A.P.

Step 1: Find the sum of the first n terms for each A.P.
Let's assume the first A.P. has a first term a1 and a common difference d1, while the second A.P. has a first term a2 and a common difference d2. The sum of the first n terms for each A.P. can be calculated using the formula Sn = (n/2)(2a + (n-1)d).

Step 2: Set up the ratio equation:
The given ratio of the sums of the first n terms is (7n-5)/(5n+17). We will set up an equation using this ratio and equate it to 1, as we need to find the term at which the sums are equal.

Step 3: Solve the equation:
To solve the equation, we will cross multiply and simplify to obtain a quadratic equation. We will then solve this equation to find the value of n.

Step 4: Find the term at which the sums are equal:
Once we have the value of n, we can substitute it back into the formula for the sum of the first n terms to find the term at which the sums are equal.

Conclusion:
By following the steps outlined above, we can find the term at which the sum of the first n terms is equal for two given arithmetic progressions.

n=117×11-5/5×11+17=72/72=1