Given information:
- The sum of the first 7 terms of an arithmetic progression (A.P.) is 49.
- The sum of the first 17 terms of the same A.P. is 289.

To find:
The sum of the first n terms of the A.P.

Solution:

Step 1: Finding the common difference (d)
In an arithmetic progression, the difference between any two consecutive terms is constant. Let's assume this common difference as 'd'.

Step 2: Finding the first term (a)
We know that the sum of the first 7 terms is 49. Using the formula for the sum of an A.P., we can express this as:
49 = (7/2)(2a + (7-1)d)
Simplifying the equation:
49 = (7/2)(2a + 6d)
98 = 2a + 6d
2a = 98 - 6d
a = 49 - 3d

Step 3: Finding the sum of the first 17 terms
We know that the sum of the first 17 terms is 289. Again, using the formula for the sum of an A.P., we can express this as:
289 = (17/2)(2a + (17-1)d)
Simplifying the equation:
289 = (17/2)(2a + 16d)
578 = 2a + 16d
Using the value of 'a' from Step 2:
578 = 2(49 - 3d) + 16d
578 = 98 - 6d + 16d
578 = 98 + 10d
10d = 578 - 98
10d = 480
d = 480/10
d = 48

Step 4: Finding the sum of the first n terms
Now that we have found the common difference, we can find the sum of the first n terms using the formula:
Sn = (n/2)(2a + (n-1)d)
Substituting the values of 'a' and 'd' from Step 2 and Step 3 respectively:
Sn = (n/2)(2(49 - 3(48)) + (n-1)(48))
Sn = (n/2)(2(49 - 144) + (n-1)(48))
Sn = (n/2)(2(-95) + (n-1)(48))
Sn = (n/2)(-190 + 48n - 48)
Sn = (n/2)(-238 + 48n)
Sn = (n/2)(48n - 238)

Therefore, the sum of the first n terms of the arithmetic progression is (n/2)(48n - 238).

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