Question

An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then the fourth term is

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

The correct option is A

2

Find the fourth term of series:

Formula:

Sum of $n$ terms of AP $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$, where $a$ is the first term and $d$ is the common difference.

Here, $S_{11}=33$

$\begin{array}{cc}\Rightarrow & 33=\frac{11}{2}\left(2a+\left(11-1\right)d\right)\\ \Rightarrow & 66=11\left(2a+11d-d\right)\\ \Rightarrow & 66=11(2a+10d)\\ \Rightarrow & 6=2a+10d\\ \Rightarrow & 3=a+5d......\left(1\right)\end{array}$

As per the given condition, every alternate term is an integer and the sum is positive.
So $a+3d=2......\left(2\right)$

On solving equation $\left(1\right)\text{and}\left(2\right)$, we get

$$\begin{gathered} \Rightarrow 2d=1 \\ \therefore d=1 \end{gathered}$$

Substitute $d=1$ in equation $\left(2\right)$

$\begin{array}{cc}\Rightarrow & a+3\left(\frac{1}{2}\right)=2\\ \Rightarrow & a=\frac{1}{2}\end{array}$

Therefore, the fourth term of AP is,

$\begin{array}{rcl}& \Rightarrow & a_4=a+3d\\ & =& \frac{1}{2}+3\left(\frac{1}{2}\right)\\ & =& 2\end{array}$

Hence, the fourth term of is $2$.