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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+(n1)d
Sum of n terms of an A.P., S_{n}=n/2[2a+(n1)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.
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