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 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 eldest participant in the painting competition.

Answer 1

This will form an AP in which,
a=8
d=4/12=1/3
Sn=168

Sn=n/2(2a+(n-1)d)
168=n/2(2(8)+(n-1)1/3)
336=n(48+n-1)/3)
$n^{2}$+47n-1008

D=$b^{2}$-4ac
 =$47^{2}-4(1)(-1008)$
  =2209+4032
  =6241

n=-b+/-√D/2a


n=-47+√6241/2=-47+79/2=32/2=16
or
n=-47-√6241/2=-47-79/2=-126/2=-63(This cannot be possible as n is the no. of students which cannot be negative)

Therefore n= 16
Tn=a+(n-1)d=8+(16-1)1/3=8+5=13 yrs

Therefore the age of the oldest person is 13 years

Hope it helped you!!  :)

Answer 2

Answer:

Step-by-step explanation:

a=8 (given)

d=4 months(given)=4\12=1\3years

sum of ages of all the participants is 168 years so

sn=n\2(2a +(n-1)d)

168=n\2(2*8 +(n-1)1\3)

by solving we get                      

n^2+4n-1008=0             also by quadratic eq "discriminant=b^2-4ac=0

                                                                                                 =  64-1008

                                                                                                 =6241