Question

Six numbers are in AP such that their sum is $3.$ The first term is $4$ times the third term. Then, the fifth term is

• A. -15
• B. -3
• C. 9
• D. -4

Answer 1

Answer: (D)

Let the numbers be
$a-5d,a-3d,a-d,a+d,a+3d,a+5d$
$\therefore a-5d+a-3d+a-d+a+d+a$
$+3d+a+5d=3$
$(\because$ Sum $=3)$
$\Rightarrow 6a=3\Rightarrow a=\frac{1}{2}$
Also, given $T_1=4T_3$, where $T_1,T_3$ are respectively, first and third terms of AP.
$\Rightarrow a-5d=4(a-d)$
$\Rightarrow d=-3a=-\frac{3}{2}$
$\therefore$ The fifth term
$a+3d=\frac{1}{2}+3\left(-\frac{3}{2}\right)=\frac{1}{2}-\frac{9}{2}=-4$