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 what?

Answer 1

Step 1: Calculate the sixth term.

Assume that the first term is $a$and the common difference is $d$.

Know that the sum of $n$terms of an A.P is $\frac{\mathbf{n}}{\mathbf{2}}\mathbf{[}\mathbf{2}\mathbf{a}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{d}\mathbf{]}$.

The sum of the first $11$terms is $33$.

Therefore,

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

Step 2: Calculate the fourth term.

Understand that here AP starts with a positive fraction and every alternate term is an integer.

Therefore,

$\begin{array}{c}2=a+(4-1)d\\ \Rightarrow 2=a+3d...........\left(2\right)\end{array}$

Step 3: Calculate the common difference.

Equate equations $\left(1\right)$and $\left(2\right)$ to calculate the common difference.

Therefore,

$\begin{array}{c}3-5d=2-3d\\ \Rightarrow 5d-3d=3-2\\ \Rightarrow 2d=1\\ \Rightarrow d=\frac{1}{2}\end{array}$

Step4: Calculate the first term.

Apply$d=\frac{1}{2}$ in equation $\left(2\right)$.

Therefore,

$\begin{array}{c}2=a+3\times \frac{1}{2}\\ \Rightarrow a=2-\frac{3}{2}\\ \Rightarrow a=\frac{4-3}{2}\\ \Rightarrow a=\frac{1}{2}\end{array}$

Step 5: Calculate the fourth term.

Therefore, the fourth term is given as:

$\begin{array}{c}{a}_4=\frac{1}{2}+(4-1)\frac{1}{2}\\ \Rightarrow a_4=\frac{1}{2}+3\times \frac{1}{2}\\ \Rightarrow a_4=\frac{1}{2}+\frac{3}{2}\\ \Rightarrow a_4=\frac{4}{2}\\ \Rightarrow a_4=2\end{array}$

Hence, the fourth term of the A.P is $2$.