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Question

Four consecutive terms in an A.P. whose sum is 12 and the sum of 3rd and 4th terms is 14 are

A
-3, 1, 5, 9
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B
3, 5, 7, 9
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C
-3, -2, -1, 0
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D
1, 5, 9, 13
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Solution

The correct option is A -3, 1, 5, 9
Let the terms be (a3d),(ad),(a+d),(a+3d). Then,
(a3d)+(ad)+(a+d)+(a+3d)=12
4a=12a=3
Also, Sum of 3rd and 4th=14
(a+d)+(a+3d)=142a+4d=14
4d=142×3d=2
Therefore, the four terms are 3,1,5,9

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