Let a be the first term and d be the common difference of the given AP. Then, the sum of first n terms is given by
`S_(n) = (n)/(2) * {2a + (n-1)d}`
`therefore S_(10) = (10)/(2) * (2a +9d) rArr 5(2a +9d) = 210`
`rArr 2a +9d = 42. " "...(i)`
Sum of last 15 terms `=(S_(50) - S_(35))`
`therefore (S_(50) - S_(35)) = 2565`
`rArr (50)/(2) (2a +49d) - (35)/(2) (2a +34d) = 2565`
`rArr 25(2a +49d)-35(a+17d) = 2565`
`rArr (50a-35a) + (1225d -595d) = 2565`
`rArr 15a +630d = 2565 rArr a +42d = 171. " "...(ii)`
On solving (i) and (ii), we get a = 3 and d = 4.
Hence, the required AP is 3, 7, 11, 15, 19,...
