Let a is the first term and d is the common difference of an A.P.

Now, nth term of an A.P. is

a_n = a + ( n – 1)d

and sum of n terms of an A.P.

S_n = (n/2){2a + (n – 1)d}

Given that the sum of the first 10 terms is 210

=> (10/2){2a + 9d} = 210

=> 5{2a + 9d} = 210

=> 2a + 9d = 210/5

=> 2a + 9d = 42  ..............1

Again, 15th term from the last = ( 50 – 15 + 1 )th = 36th term from the beginning

Now, a₃₆ = a + 35d

Sum of the last 15 terms = (15/2) [2a₃₆ + ( 15 – 1)d ] = 2565

=> (15/2){2(a + 35d) + 14d} = 2565

=> 15(a + 35d + 7d) = 2565

=> a + 42d = 2565/15

=> a + 42d = 171  ..............2

Solve equation 1 and 2, we get

d = 4 and a = 3

Therefore, the terms in A.P. are: a, a + d, a + 2d, a + 3d, .........., a + 49d 

i.e. 3, 7, 11, 15, ........., 199