Question

Six numbers are in an 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

The correct option is D

$-4$

Explanation for the correct option:

Step 1: Finding the values of $a$ and $d$

Let $a-5d,a-3d,a-d,a+d,a+3d,a+5d$ be the numbers in AP.

Sum of the given A.P $=3$. Therefore,

$$\begin{gathered} a-5d+a-3d+a-d+a+da+3d+a+5d=3 \\ 6a=3 \\ a=\frac{1}{2} \end{gathered}$$

Given the first term is $4$times the third term.

$T_1=4T_3$

where$T_1=a-5d\text{and}{\text{T}}_3\text{=a-d}$

Thus,

$$\begin{gathered} a-5d=4(a-d) \\ \Rightarrow 3a+d=0 \\ \Rightarrow \frac{3}{2}+d=0 \\ \Rightarrow d=-\frac{3}{2} \end{gathered}$$

Step 2: Finding the fifth term of the AP

The fifth term is $=a+3d$

$$\begin{gathered} =\left(\frac{1}{2}\right)+3\left(-\frac{3}{2}\right) \\ =\frac{1}{2}-\frac{9}{2} \\ =-\frac{8}{2} \\ =-4 \end{gathered}$$

Thus, $a+3d=-4$

Therefore, option (D) is the correct answer.