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Question
The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the 6th and the 10th terms of the same A.P. is 34. Find the first three terms of the A.P.
Solution
Let 'a' be the first term and 'd' be the common difference of the given A.P.
t4 + t8 = 24 ...(Given)
`=>` (a + 3d) + (a + 7d) = 24
`=>` 2a + 10d = 24
`=>` a + 5d = 12 ...(i)
And t6 + t10 = 34 ...(Given)
`=>` (a + 5d) + (a + 9d) = 34
`=>` 2a + 14d = 34
`=>` a + 7d = 17 ...(ii)
Subtracting (i) from (ii), we get
2d = 5
`=> d = 5/2`
`=> a + 5 xx 5/2 = 12`
`=> a + 25/2 = 12`
Thus we have,
1st term = `-1/2`
2nd = a + d
= `-1/2 + 5/2`
= 2
3rd term = a + 2d
= `-1/2 + 2 xx 5/2`
= `-1/2 + 5`
= `9/2`
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