Question

In an A.P. of 50 terms, the sum of first 10 terms is 250 and the sum of its last 15 terms is 2625. Find the A.P. so formed.

Answer 1

Let a and d be the first term and the common difference of an A.P., respectively.

n = 50

Given, a sum of the first 10 terms = 250

`S_n = n/2[2a + (n - 1)d]`

`S_10 = 10/2[2a + (10 - 1)d]`

250 = `10/2[2a + 9d]`

250 = 5 [2a + 9d]

2a + 9d = 50     ...(1)

15^th term from the last = (50 − 15 + 1)^th = 36^th term from the beginning.

36^th term = a + 35d

Now, the sum of the last 15 terms is 2625.

`S_("last 15") = 15/2[2(a + 35d) + (15 - 1)d] = 2625`

`15/2[2(a + 35d) + 14d] = 2625`

`15/2[2a + 70d + 14d] = 2625`

`15/2[2a + 84d] = 2625`

15[2a + 84d] = 5250

2a + 84d = 350     ...(2)

Subtract Equation (2) from Equation (1), we get

2a + 9d = 50
2a + 84d = 350
-      -        -        
       -75d = -300

d = `300/75`

d = 4

By putting the value of d in Equation (1), we get

⇒ 2a + 9 × 4 = 50

⇒ 2a + 36 = 50

⇒ 2a = 50 − 36

⇒ 2a = 14

⇒ a = 7

The first term a = 7 and the common difference d = 4.
Thus, the A.P. is a, a + d, a + 2d, a + 3d, ......, a +49d

∴ A.P. is 7, 11, 15, 19, ...., 203