Question

The average of 16 consecutive positive integers is 30.5. What is the average of the first 8 integers of this set?

Answer 1

consecutive integers are 1 apart

16 of them can be represented as

x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9, x+10, x+11, x+12, x+13, x+14, x+15

the average is $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

there might be some trick to get the average of the first 8 integers only

the average of the first 8 integers would be $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$

how can we get $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$ from $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

first, let's match the denomenators (make them both 1 for ease)

multiply the first eqution by 8 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7=8?

multiply the 2nd equation by 16 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

notice that we can try and force the 1st equation into the 2nd equation by adding some numbers

we can already do this:

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[8?]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

we can try to force another 8? into it

[8?]+x+7+1+x+7+2+x+7+3+x+7+4+x+7+5+x+7+6+x+7+7+x+7+8=488

[8?]+(x)+7+(1+x)+7+(2+x)+7+(3+x)+7+(4+x)+7+(5+x)+7+(6+x)+7+(7+x)+7+8=488

[8?]+(x)+7+(x+1)+7+(x+2)+7+(x+3)+7+(x+4)+7+(x+5)+7+(x+6)+7+(x+7)+7+8=488

group

[8?]+[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+7+7+7+7+7+7+7+7+8=488

[8?]+[8?]+8(7)+8=488

16?+64=488

minus 64 from both sides

16?=424

divide both sides by 16

?=26.5

the average of the first 8 integers is 26.5

I'm not sure if there is a simpler way to do it or not