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 (a−3d),(a−d),(a+d),(a+3d). Then, (a−3d)+(a−d)+(a+d)+(a+3d)=12 ⇒4a=12⇒a=3 Also, Sum of 3rd and 4th=14 ⇒(a+d)+(a+3d)=14⇒2a+4d=14 ⇒4d=14−2×3⇒d=2 Therefore, the four terms are −3,1,5,9