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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 sixth term and the tenth is 44. Find the first three terms of the A.P.
Sum
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Solution
Given
t4 + t8 = 24
`=>` (a + 3d) + (a + 7d) = 24
`=>` 2a + 10d = 24
`=>` a + 5d = 12 ...(i)
And,
t6 + t10 = 44
`=>` (a + 5d) + (a + 9d) = 44
`=>` 2a + 14d = 44
`=>` a + 7d = 22 ...(ii)
Subtracting (i) from (ii), we get
2d = 10
`=>` d = 5
Substituting value of d in (i), we get
a + 5 × 5 = 12
`=>` a + 25 = 12
`=>` a = -13 = 1st term
a + d = -13 + 5 = -8 = 2nd term
a + 2d = -13 + 2 × 5 = -13 + 10 = -3 = 3rd term
Hence, the first three terms of an A.P are -13, -8 and -3
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