Question

in an AP 50 terms the sumof first 10 terms is 210 and the sum of its last 15 terms is 2565. find the AP.

Answer 1

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an = a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36 = a + 35d

Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

Answer 2

Let a be the first term and d be the common difference of the given AP.
Sum of the first n terms is given by
Sn = n/2 {2a + (n - 1)d}
Putting n = 10, we get
S₁₀ = 10/2 {2a + (10 - 1)d}
210 = 5 (2a + 9d) 
2a + 9d = 210/5
2a + 9d = 42 ...............(1)
Sum of the last 15 terms is 2565
⇒ Sum of the first 50 terms - sum of the first 35 terms = 2565
S₅₀ - S₃₅ = 2565
⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 2565
25 (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 get
2a + 84d = 342 .........(3)
Subtracting (1) from (3)
  2a + 84d = 342
  2a + 9d   =  42
-      -         -
_______________
        75d = 300
_______________
 d= 4
Now, substituting the value of d in equation (1)
2a + 9d = 42
2a + 9*4 = 42
2a = 42 - 36
2a = 6
a = 3
So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ..