A class consists of a number of boys whose ages are in A.P. The common difference being four months. If the youngest boys is $8$ years and if the sum of ages is $168$ years then find the number of boys.

Given: A class consists of a number of boys whose ages are in A.P. The common difference being four months. If the youngest boys is $8$ years and if the sum of ages is $168$ years.

To do: To find the number of boys.

Solution:

Here, First term$=$age of youngest boy, $a=8$

Here, $a=8,\ d=4\ months=\frac{4}{12}\ years=\frac{1}{3}\ years$

$S_n=\frac{n}{2}[2a+(n-1)d]=168$

$\Rightarrow n.\frac{2\times8+( n-1)\frac{1}{3}}{2}=168$

$\Rightarrow n.\frac{48+( n-1)1}{2.3}=168$

$\Rightarrow 48n+n^2-n=6\times168$

$\Rightarrow n^2+47n-1008=0$

$\Rightarrow ( n-16)( n+63)=0$

$n=16\ or\ -63$

$\because n≥0$, because it is no. of persons.

$\therefore n=16$

Thus, the no. of persons is $16$.

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