Question

The ages of the boys in a group in A.P.with the common difference of 3 months.The age of the youngest boy is 12 years.The sum of the age of all boys in group is 375 years .Find the numbers of boys in the group

Answer 1

Solution :-

Let the number of boys in the group be 'n'

Given - First term 'a' = 12 (Age of youngest boy is 12 years)
 
Common difference = 3 months = 3/12 years = 1/4 years 

Sn = 375 years

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

375 = n/2[2*12 + {(n - 1)1/4}]

375*2 = n[24 + {(n - 1)/4}]

750 = [{24n/1} + {(n² - n)/4}]

750 = (96n + n² - n)/4

750*4 = n² + 95n 

n² + 95n - 3000 = 0

n² + 120 - 25n = 3000 = 0

n(n + 120) - 25(n + 120) = 0

(n + 120) = 0 or (n - 25) = 0

n = - 120 or n = 25

Number of boys cannot be negative so, the correct value of n is 25.

Hence, total number of boys in the group is 25

Answer.

Answer 2

Answer:

Here, the total sum of ages i. e. Sn=375years

= 375 x12 = 4500months (1 year =12months )

First term (a) = 12 years. i. e 144months

Common difference(d)= 3months

Total number of boys (n)=?

ACCORDING TO FORMULA

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

=4500=n/2(2x144 +(n-1)3)

=4500x2=n(288+3n-3)

=9000=n(285+3n)

=9000=285n+3n^2

=3n^2+285n-9000=0

Divide throughout by 3

=n^2 +95n-3000

Solve this quadratic equation through factorisation method

The factors are -120and 25

(Note:- This quadratic equation can also be solved through completing square method or formula method)

=n^2 -120n+25n-3000

=n(n+120)-25(n+120)

=n+120 ; n-25

=n=-120 ; n=25

But the number of boys can never be negative

•: n=25