Question

In an ap 50 terms are there the sum if first 10 terms is 210and the sum of last 15terms is 2565 find an ap​

Answer 1

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 ........

hope it helps you dear

Answer 2

let say ap

a , a+d , a +2d, a +48d , a + 49d

sum of n terms = (n/2)(first term + last term)

sum of first 10 terms = (10/2)(a + a + 9d)

5 (2a + 9d) = 210

2a + 9d = 42 eq1

last 15 terms means terms from 36 to 50

36th term = a + 35d

50th term = a + 49d

now we can say 36th term is first term and n = 15

sum of last 15 terms = (15/2)(a+35d + a +49d)

(15/2)(2a + 84d) = 2565

2a + 84 d = 342 eq2

eq2 -eq1

75d = 300

d = 4

2a + 9d = 42

2a + 36 = 42

2a = 6

a = 3

AP

3 , 7 ,11 ...............,195 , 199