Question

5

21. The average of 5 consecutive integers starting with m
as the first integer is n. What is the average of 9
consecutive integers that start with m+2?
A) m + 4
B) n + 6
C) n+ 4
D) m + 5

Answer 1

n + 4

Explanation:

Given

⅕(m + m + 1 + m + 2 + m + 3 + m + 4) = n

Required

⅑(m + 2 + m + 3 +.......+ m + 10)

We have:

⅕(m + m + 1 + m + 2 + m + 3 + m + 4) = n

Multiply both sides by 5

m + m + 1 + m + 2 + m + 3 + m + 4 = 5n

Collect like terms

m + m + m + m + m = 5n - 1 - 2 - 3 - 4

5m = 5n - 10

Divide both sides by 5

m = n - 2

So, we have:

⅑(m + 2 + m + 3 + m + 4 + m + 5 + m + 6 + m + 7 + m + 8 + m + 9 + m + 10)

Collect like terms

= ⅑(m+m+m+m+m+m+m+m+m+2+3+4+5+6+7+8+9+10)

= ⅑(9m + 54)

= m + 6

Substitute n - 2 for m

= n - 2 + 6

= n + 4