Question

In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Answer 1

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

n^th term of an A.P. a_n=a+(n-1)d

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

We have:
Sum of the first 10 terms =10/2[2a+9d]

210=5[2a+9d]

42=2a+9d            ...........(1)

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

Now, a₃₆=a+35d 

∴Sum of the last 15 terms `=15/2(2a_36+(15−1)d)  `

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

`= 1/5[a+35d+7d]`

`2565=15[a+42d]`

`171=a+42d .................(2)`

From (1) and (2), we get:

d = 4
a = 3

So, the A.P. formed is 3, 7, 11, 15 . . . and 199.