Question

6. The ages of the students of a class are in A.P., whose common difference is $4$months. If the youngest boy is $8$ years old and the sum of the ages of all the students of the class is $168$ years. Find the numbers of students in the class.

Answer 1

Step 1. Note the given data.

Given : Age of youngest boy $=8years$

Common difference in ages of students$=4months$

Sum of ages of all the students of the class$=168years$

Let the number of students be $n$.

Step 2. Find the first term, common difference, and sum in months.

Let, First term is$a$

$$\begin{gathered} a=8years \\ =8\times 12months \\ =96months. \end{gathered}$$

Common difference $d=4months$

And sum,

$$\begin{gathered} S_n=168years \\ =168\times 12months \\ =2016months \end{gathered}$$

Step 3. Find the number of students.

The formula for the sum of $n$ terms in A.P is- $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

The sum of $n$ terms of the AP is $S_n=\frac{n}{2}\left[2\times 96+\left(n-1\right)4\right]$

$$\begin{gathered} =\frac{n}{2}\left[192+4n-4\right] \\ =\frac{n}{2}\left[188+4n\right] \\ =\frac{n}{2}\times 2\left[94+2n\right] \\ =94n+2n^2 \end{gathered}$$

Step 4. Solve using a quadratic equation.

Now, we put the values as-

$$\begin{gathered} 94n+2n^2=2016 \\ 2n^2+94n-2016=0 \\ {n}^2+47n-1008=0 \\ {n}^2+63n-16n-1008=0 \end{gathered}$$

$$\begin{gathered} n(n+63)-16(n+63)=0 \\ (n+63)(n-16)=0 \\ Eitheror, \\ n+63=0n-16=0 \\ n=-63n=16 \end{gathered}$$

$n$shouldn't be negative.

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