Question

the averages of 5 consecutive integers starting with m as the first integer is n. what is the average of following integers?
m+1,m+3,m+5,m+7,m+9

Answer 1

[(m+1)+ (m+3)+ (m+5) + (m+7) + (m+9) ] / 5

Answer 2

next step?

Answer 3

then collect like terms

Answer 4

those are consecutive odds and evens.

consecutive alone means 1,2,3,4,5.. (one right after the other)
so here it'd be \(\large \frac{m+(m+1)+(m+2)+(m+3)+(m+4)}{5}=n\)

they're asking for m+1,m+3,m+5,m+7,m+9 (which simplifies to 5m+25)

our original fraction says m+m+1+m+2+m+3+m+4=5m+10

so we have the fact that \(\large \frac{5m+10}{5}=n\) and \(\large \frac{5m+25}{5}=n+\frac{15}{5}\)

Answer 5

optiona are
m+4
n+6
m+5

Answer 6

@bibby

Answer 7

@perl

Answer 8

m+5? is ans i guess

Answer 9

( m + 25 ) / 5 = m/5 + 5