If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP.

In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n - 1)d ​
Now, we have:
T10 = a + (10 - 1)d 
⇒ a + 9d  = 52       ...(1)
T13 = a + (13 - 1)d = a + 12d               ...(2)
T17 = a + (17 - 1)d = a + 16d                 ...(3)     

But, it is given that T17 = 20 + T13
i.e., a + 16d  = 20 + a + 12d 
⇒ 4d = 20  
⇒ d = 5
On substituting d = 5 in (1), we get:
a + 9 ⨯ 5 = 52 
⇒​ a = 7