Saved Bookmarks
| 1. |
In an AP of 50terms, the sum of 1st 10 terms is 210 and the sum of last15 terms is 2565.finr AP |
|
Answer» AP of an=a+ ( n-1)d, sum of sn=n/2(2a+( n-1)d, sum first 10 terms is 210. 10/2(2a+9d)=210, 5(2a+9d)=210, ==2a+9d=42__________(1), 15 terms the last=(50-15+1)th=35d; 15/2(2a36+(15-1)d)=2565; 15/2(2a+35d)+14d)=2565; 15(a+35d+7d)=2565; a + 42d=171_______(2), from (1) and (2)= d/4; a=3, A.P. are 3, 7, 11, 15 _____,,,,199 S₅₀ - S₃₅ = 2565⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 256525 (2a + 49d) - 35/2 (2a + 34d) = 2565⇒ 5 (2a + 49d) - 7/2 (2a + 34d) = 513⇒ 10a + 245d - 7a + 119d = 513⇒ 3a + 126d = 513⇒ a + 42d = 171 ........(2)Multiply the equation (2) with 2, we get2a + 84d = 342 .........(3)Subtracting (1) from (3) 2a + 84d = 342 2a + 9d = 42- - -_______________ 75d = 300_______________d= 4Now, substituting the value of d in equation (1)2a + 9d = 422a + 9*4 = 422a = 42 - 362a = 6a = 3So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ........ |
|