In an ap of 50terms the …

Let a be the first term and d be the common difference of the given AP. Therefore, the sum of first n terms is given by
{tex}S _ { n } = \frac { n } { 2 } \cdot \{ 2 a + ( n - 1 ) d \}{/tex}
{tex}\therefore{/tex} S₁₀ = {tex}\frac{{10}}{2}{/tex}{tex}\cdot{/tex}(2a+9d) {tex}\Rightarrow{/tex}5(2a+9d)=210
{tex}\Rightarrow{/tex}2a+9d=42. ...(i)
Sum of last 15 terms = (S₅₀ - S₃₅).
{tex}\therefore{/tex} (S₅₀ - S₃₅) = 2565
{tex}\Rightarrow{/tex} {tex}\frac{{50}}{2}{/tex}(2a+49d)- {tex}\frac{{35}}{2}{/tex}(2a+34d)=2565
{tex}\Rightarrow{/tex} 25(2a+49d)-35(a+17d)=2565
{tex}\Rightarrow{/tex} (50a-35a)+(1225d-595d)=2565
{tex}\Rightarrow{/tex} 15a+630d = 2565 {tex}\Rightarrow{/tex} a + 42d = 171 ...... (ii)
Therefore, on solving (i) and (ii), we get a=3 and d=4.
Hence, the required AP is 3,7,11,15,19.....