In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Let a be the first term and d be the common difference of the given A.P 
We know that sum of first n terms of an A.P is given by,
$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}} = \frac{\mathrm{n}}{2}\left[2\mathrm{a} + \left(\mathrm{n} - 1\right) \mathrm{d}\right] \\ \mathrm{Put} \mathrm{n} = 10, \mathrm{we} \mathrm{get} \\ {\mathrm{S}}_{10} = \frac{10}{2}\left[2\mathrm{a} + 9\mathrm{d}\right] \\ \Rightarrow 210 = 5 \left(2\mathrm{a} + 9\mathrm{d}\right) \\ \Rightarrow 2\mathrm{a} + 9\mathrm{d} = 42 ...........\left(1\right) \\ \mathrm{Sum} \mathrm{of} \mathrm{last} 15 \mathrm{terms} = 2565 \\ \Rightarrow \mathrm{Sum} \mathrm{of} \mathrm{first} 50 \mathrm{terms} - \mathrm{sum} \mathrm{of} \mathrm{first} 35 \mathrm{terms} = 2565 \\ \Rightarrow {\mathrm{S}}_{50} - {\mathrm{S}}_{35} = 2565 \\ \Rightarrow \frac{50}{2}\left[2\mathrm{a} + (50 - 1)\mathrm{d}\right] - \frac{35}{2}\left[2\mathrm{a} + (35 - 1) \mathrm{d}\right] = 2565 \\ \Rightarrow 25(2\mathrm{a} + 49\mathrm{d}) - \frac{35}{2}\left(2\mathrm{a} + 34\mathrm{d}\right) = 2565 \\ \Rightarrow 5(2\mathrm{a} + 49\mathrm{d}) - \frac{7}{2}\left(2\mathrm{a} + 34\mathrm{d}\right) = 513 \\ \Rightarrow 3\mathrm{a} + 126\mathrm{d} = 513 \\ \Rightarrow \mathrm{a} + 42\mathrm{d} = 171 ...........\left(2\right) \\ \mathrm{multiply} \left(2\right) \mathrm{with} \text{'}2\text{'} \mathrm{we} \mathrm{get} \\ 2\mathrm{a} + 84\mathrm{d} = 342 ..........\left(3\right) \\ \mathrm{subtracting} \left(1\right) \mathrm{from} \left(3\right), \mathrm{we} \mathrm{get} \\ 84\mathrm{d} - 9\mathrm{d} = 342 - 42 \\ \Rightarrow 75\mathrm{d} = 300 \\ \Rightarrow \mathrm{d} = \frac{300}{75} = 4 \\ \mathrm{now}, \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ \mathrm{a} +42\left(4\right) =171 \\ \mathrm{a} + 168=171 \\ \mathrm{a} = 3 \end{gathered}$$

so, the AP is a, a+d, a+2d, a+3d, .........
so the required AP is, 3, 7, 11, 15, .......