Question

The minimum age of children to be eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of rest of participants are having a common difference of 4 months. If the sum of ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.

Answer 1

Hint: The ages of all the participants in the painting competition are in AP. The first term of AP is 8 and the common difference is 4 months. Now, sum of all the participants ages = 168, we use the formula for sum of all terms in AP as follows ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, we get the value of ‘n’ and then we will find ${{t}_{n}}$ as $a+\left( n-1 \right)d$.

Complete step-by-step solution -
It is given in the question that the minimum age of children to be eligible to participate in painting competition is 8 years. Also, it is observed that the youngest boy among all participants, which is of 8 years and there is a pattern, common difference of 4 months in ages of all the participants. Also, the sum of ages of all the participants = 168.
It means the sum of all the participants are in AP having first term = 8 and the common difference is 4 months. We know that sum of all the n terms in AP is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ‘a’ is the first term, ‘d’ is the common difference, ‘n’ is the number of terms. Therefore putting the values in the given formula, we get $168\times 12=\dfrac{n}{2}\left[ 2\times 8\times 12+\left( n-1 \right)4 \right]$ simplifying further, we get $2016=\dfrac{n}{2}\left[ 192+4n-4 \right]$ on solving further, we get $4032=n\left[ 188+4n \right]$.
Therefore, we get a quadratic equation as $4032=188n+4{{n}^{2}}$ or $4{{n}^{2}}+188n-4032=0$, taking 4 common from LHS and transposing it to RHS we get ${{n}^{2}}+47n-1008=0$. Now solving this quadratic equation we find the values of ‘n’ as the roots of this equation. That is $n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$n=\dfrac{-47\pm \sqrt{{{47}^{2}}-4\times 1\times \left( -1008 \right)}}{2\times 1}$
$n=\dfrac{-47\pm \sqrt{2209+4032}}{2}$
$n=\dfrac{-47\pm 79}{2}$ from here we get two values of ‘n’ as $n=\dfrac{-47+79}{2}$ and $n=\dfrac{-47-79}{2}$. Since the number of terms cannot be negative thus we neglect the second value, that is, negative value of n. Thus we get $n=\dfrac{-47+79}{2}=\dfrac{32}{2}=16$. Thus $n=16$, which means the 16th participant in the competition has the maximum age.
Now, we know that ${{t}_{n}}=a+\left( n-1 \right)d$, therefore ${{t}_{16}}=8+\left( 16-1 \right)\dfrac{1}{3}$ simplifying further, we get ${{t}_{16}}=8+15\times \dfrac{1}{3}=8+5=13$ years. Therefore, the age of the eldest participant will be 13 years.

Note: Usually student skip to convert years into month in the first step and take 4 as the common difference but if we are taking 4 as the common difference then we have to convert all the given age in years into months by multiplying by 12 or if we are taking age in years then the common difference will be $\dfrac{4}{12}=\dfrac{1}{3}$ years.