Question

The ages of the students in a class are in AP, whose common difference is $4$months. If the youngest student is $8yr$ old and the sum of the ages of all the students is $168yr$ then find the number of students in the class.

Answer 1

Step 1. Given:

Here, the age of the youngest student, $a=8yr$.

Common difference,

$\begin{array}{rcl}d& =& 4\text{months}\\ & =& \frac{4}{12}\text{yr}\\ & =& \frac{1}{3}\text{yr}\end{array}$.

The sum of the ages of all the students in the class is $168yr$.

$\therefore S_n=168$

Let $n$ be the number of students in the class.

Step 2. Find the quadratic equation by using the formula $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$:

Formula:

We know that the sum of the first $n$ terms of an AP is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Where $a=$ first term and $d=$ common difference.

$$\begin{gathered} \Rightarrow 168=\frac{n}{2}\left[2\times 8+(n-1)\times \frac{1}{3}\right] \\ \Rightarrow 168=\frac{n}{2}\left[16+(n-1)\times \frac{1}{3}\right] \\ \Rightarrow 168=\frac{n}{6}\left[48+(n-1)\right] \end{gathered}$$

$\Rightarrow 48n+n^2-n=1008$

Step 2. Solve the quadratic equation to find the values of $n$:

$$\begin{gathered} \begin{array}{cc}\Rightarrow & n^2+47n-1008=0\\ \Rightarrow & n^2+63n-16n-1008=0\\ \Rightarrow & n(n+63)-16(n+63)=0\\ \Rightarrow & (n+63)\left(n-16\right)=0\end{array} \\ \Rightarrow n=16\mathrm{and}n=-63 \end{gathered}$$

$\because$Number of students cannot be negative

Hence, the number of students in the class is $16$.