Question

the minimum age of children to be eligible to participate ina painting competition is 8 years . it is observed thatbthe age of youngest boy was 8 years and the ages of rest of participants are having a common difference of 4 monts. of tge sum of ages of all the participants is 168 years, find age of eldwst participate in the competition ​

Answer 1

HeYa❤️...

Answer:

Here, a=8 , d= 4 months = 1/3 years and Sn= 168

Sn=n/2[2a+(n-1)d]

$168 = \frac{n}{2}(2 \times 8 + (n - 1) \frac{1}{3} \\ \\ n {}^{2} + 47n - 1008 = 0 \\ \\ n {}^{2} + 63n - 16n - 1008 = 0 \\ \\ (n - 16)(n + 63) = 0 \\ \\ n = 16 \: or \: n = - 63 \\ \\ n = 16$

Thus, the age of eldest participant= a + 15d = 13 years.

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Answer 2

Answer:

Step-by-step explanation:

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)

+47n-1008

D=-4ac

 =

 =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