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.

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#### Solution

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, a_{36}=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.

Concept: Arithmetic Progression

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