Aclass consists of a number of boys whose ages are in ap, the common
difference four months. if the youngest boy is just eight years old, and if
the sum of ages is 168 years, find the number of boys in the class.

n = 16

Step-by-step explanation:

Formula for finding the sum of an AP

Sₙ = ⁿ/₂ {(2a + ( n - 1 )d )}

where a = 8, d is 4 months = ¹/₃ of a year and ₎ and Sₙ = 168

Therefore, substitute for the values an form an equation with it

168 = ⁿ/₂{( 2(8) + ( n - 1 )¹/₃}

multiply by 2 to make it a linear expression

336 = n( 16 + ⁿ/₃ - ¹/₃ )

336 = n(  16 - ¹/₃ + ⁿ/₃ )

336 = n( 48 - 1 /3 + ⁿ/₃

Now open the brackets

336 = n(⁴⁷/₃ + ⁿ/₃ )

336 = ⁴⁷ⁿ/₃ + n²/₃

Multiply through by 3

1,008 = 47n + n², then rearrange

n² + 47n - 1008 = 0

n² + 63n - 16n - 1008 = 0

factorize the expression by grouping

n(n + 63) - 16(n + 63) = 0

pick common factors

(n + 63)(n - 16) = 0

therefore n = -63 or 16. But note that n ≠ -63,

Therefore n = 16 which is the required number of boys in the class.

1 is c

2 is a

3 is a

step-by-step explanation:

2

step-by-step explanation: