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The minimum age of children 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 the participants, when seated in order of age, have a common difference of $4$ months. If the sum of the ages of all the participants is $168$ years, find the age of the eldest participant in the painting competition.
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Sol. The ages of the participants form the following AP
$8, 8\frac{1}{3}, 8\frac{2}{3}, 9, ...$
where first term $= 8$ and common difference $= \frac{1}{3}$
Let the number of participants be $n$
$S_n = \frac{n}{2}[2 \times 8 + (n-1)\frac{1}{3}] = 168$
$n^2 + 47n - 1008 = 0$
$\Rightarrow n = 16$
$\therefore$ the age of the eldest participant $= 8 + 15 \times \frac{1}{3} = 13$ years
$8, 8\frac{1}{3}, 8\frac{2}{3}, 9, ...$
where first term $= 8$ and common difference $= \frac{1}{3}$
Let the number of participants be $n$
$S_n = \frac{n}{2}[2 \times 8 + (n-1)\frac{1}{3}] = 168$
$n^2 + 47n - 1008 = 0$
$\Rightarrow n = 16$
$\therefore$ the age of the eldest participant $= 8 + 15 \times \frac{1}{3} = 13$ years