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Question
The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P
Solution
The general term of an AP is given by `t_n = a + (n - 1)d`
Now `t_2 + t_7 = 30`
=> (a + d) + (a + 6d) = 30
=> 2a + 7d = 30 ...(i)
Next `2 xx t_8 - t_15= 1`
`=> 2 xx (a + 7d) - (a + 14d) = 1`
=> 2a + 14d - a - 14d = 1
=> a = 1
Substituting the value of a in (i) we get
`2 xx 1+ 7d = 30`
=> 7d = 28
`=> d = 4`
Thus required A.P = a, a + d, a + 2d, a + 3d
= 1, 5, 9, 13, ....
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