Question

ages of boys in a group are in AP with common difference of 3 months.age of youngest boy in group is 12 yrs.Sum of ages of all boys in group is 375.Find no. of boys in the group

Answer 1

Solution
Let the number of boys in the group be
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/212'12+ (n - 1)1/4]
375*2 = n24 + {(n-1)/41
750 = [24n/1) + (n2-n)/41
750 = (96n + n2-n)/4
750*4 n2 + 95n
n2 + 95n-3000 ?
n2 + 95n-3000 = 0
+ 120-25n 3000-0
n(n 120) - 25(n + 120)0
(n + 120) O 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

given common difference d = 3 months Youngest boy = 12 years old a = 12 years = 12 × 12 =144 months Sn = 375 years = 375 × 12 = 4500 months Sn = n/2 [2a + (n - 1) d] 4500 = n/2 [(2 × 144) + (n - 1) 3] 4500 × 2 = n [288 + 3n - 3 ] 9000 = 3n^2 + 285n i.e. 9000 is equal to 3n square plus 285n Dividing by 3 on both sides n^2 + 95n - 3000 = 0 n^2 + 120n - 25n - 3000 = 0 n (n + 120) - 25 (n + 120) = 0 (n + 120) (n - 25) = 0 n = - 120 or n = 25 Total no of student never negative Hence n = 25 is the correct answer
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