Question

What is the average of all of the integers from 13 to 37?

Answer 1

sum them all up with the AP sum formula... then divide by the no. of nos. to get the mea, or the 'avg.'

Answer 2

(13+37)/2 .. if it is symmetric on some no

Answer 3

I thought @Tributized might be right .. if you can constrct (24+25+26)/3 = 25

Answer 4

i deleted it because u were teaching the asker ._. i don't wanna give the answer yet lol

Answer 5

(23+24+25+26+27)/5 = 25
(13+ ... +23+24+25+26+27+ ,,,,, 37)/(37-13) = 25

Answer 6

no .. i just came up with a hunch .. i haven't calculated yet ... this is just interesting ... to predict result without calculating

Answer 7

I see

Answer 8

oh ok. so the answer is 25. :D

Answer 9

seriously, i haven't checked it yet .... just a prediction

Answer 10

everybody is giving me medals .. guess i was right or what :D

Answer 11

\[\frac12[\sum_{i=1}^{37}i-\sum_{i=1}^{12}i]\]eh?
may not be the most efficient way

Answer 12

yep ur right. since im right :D

Answer 13

The average (or mean) is the same as the median with a series of numbers which differentiate from each other by the same number (in this case, 1)

Answer 14

(13+37)/2 = (14+36)/2 = (15+35)/2 = ... = (24+26)/2 = 25/1 = 25,

you only needed to calculate (13+37)/2 =25.

The same is true for any sum of consecutive integers.

Answer 15

Turing, why dividing by 2?

Answer 16

oops
divide by 37-12

Answer 17

right?

Answer 18

the sum is
\[\S_{n}=\frac{n}{2}(a_{1}+a_{n})\]
the average would be
\[\frac{\S_{n}}{a_{n}-a_{1}}=\frac{\frac{a_{n}-a_{1}}{2}(a_{1}+a_{n})}{a_{n}-a_{1}}=\frac{\frac{37-13}{2}(37+13)}{37-13}=\frac{37+13}{2}\]

Answer 19

13 through 37 is the same as 12(25) = 300
( there are 25 numbers to add )
add to this the numbers 1 through 25
The formula for this is:
(1/2) x 25 x 26 = 325
300 + 325 = 625
625/25 = 25

Answer 20

I did this \[ (37 \times 38/2 - 12 \times 13/2)/(37-12) = 25\]

Answer 21

yeah that's easier lol

Answer 22

oh jeez, I added in my calculator
at least that earlier formula is right then with the modification mentioned by FFM\[\frac1{25}[\sum_{i=1}^{37}i-\sum_{i=1}^{12}i]\]