AP of the 15 the sum of the first 10 terms is 210 the sum of the last 15 terms is 2565 find the AP

AP of the 15 the sum of the first 10 terms is 210 the sum of the last

1.	AP of the 15 the sum of the first 10 terms is 210 the sum of the last 15 terms is 2565 find the AP
Answer» 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}\xa0S10 = {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}\xa0(S50\xa0- S35) = 2565{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{{50}}{2}{/tex}(2a+49d)- {tex}\\frac{{35}}{2}{/tex}(2a+34d)=2565{tex}\\Rightarrow{/tex}\xa025(2a+49d)-35(a+17d)=2565{tex}\\Rightarrow{/tex}\xa0(50a-35a)+(1225d-595d)=2565{tex}\\Rightarrow{/tex}\xa015a+630d = 2565 {tex}\\Rightarrow{/tex}\xa0a + 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.....