Question

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

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

Answer 1

The correct option is A -3, 1, 5, 9

Let the terms be $(a-3d),(a-d),(a+d),(a+3d).$ Then,
$(a-3d)+(a-d)+(a+d)+(a+3d)=12$
$\Rightarrow 4a=12\Rightarrow a=3$
Also, Sum of $3^{rd}$ and $4^{th}=14$
$\Rightarrow (a+d)+(a+3d)=14\Rightarrow 2a+4d=14$
$\Rightarrow 4d=14-2\times 3\Rightarrow d=2$
Therefore, the four terms are $-3,1,5,9$