Odd Mean

Actually the best solution for this problem is not sum up $1,3,5...$

The interesting fact is:

$1^2=1$

$2^2=1+3$

$3^2=1+3+5$

and so on.

From the pattern above, immediately we can see that the sum of the first 9 odd integers is

$9^2$

For the mean, or average, we divide it by the number of terms, which is $9$ and we can get

$\frac{9^2}{9}$

$=9$

Brute force is ok either, but this is another way to approach! :)

Sum of first 9 positive odd integers is 81.

Mean = 81/9 = 9.

First 9 positive odd integers goes like this: 1, 3, 5, .... (9 values). Here we don't know the 9th value. So, to get the 9th value we see that the sequence is of the form 2n+1, where n starts from 0. To get the 9th value, n must be 8. Substitute n as 8, we get the 9th value as 17. Now since they are A.P., the average of extreme values is the average of the whole sequence, i.e. 1 + 17 divided by 2 is the average of the sequence. So the answer is 9. Just observe that 3 + 15 divided by 2 is also the mean, same with 5 + 13 divided by 2 and so on. That's the typical property of an A.P. (Arithmetic Progression).

The mean of the first 9 positive odd integer is the fifth number : 9.

There are 2 important hints that we have to look out in order to solve this riddles. The first clue is that the integers chosen from a list of numbers should be positive, so it should be starting from 1,2,3,4....... 0 is not regarded as positive integers as it is a neutral number. The second clue is that the integers that we should opt for is consist of 9 different odd numbers. By now, we should have know that the set of numbers required is 1,3,5,7,9,11,13,15 and finally 17.

Hence, Mean=(1+3+5+7+9+11+13+15+17)/9 =(81)/9 =9

Add all the numbers and divide the total by 9

Another approach to get the answer:

First nine of odd integer are $1, 3, 5, 7, 9, 11, 13, 15, 17$ . Since the number of list is odd, finding mean, initially, you can manipulate that list using middle term of them.

$\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17}{9} =$

$= \frac{(9 - 8) + (9 - 6) + (9 - 4) + (9 - 2) + 9 + (9 + 2) + (9 + 4) + (9 + 6) + (9 + 8)}{9}$

$= \frac{9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9}{9}$

$= \frac{9 * 9}{9} = 9$

It is quit simple.Mean =sum of the terms/number of terms. The first 9 positive odd integers are 1,3,5,7,9,11,13,15,17.Sum of the terms is 81.Number of terms is 9. Thus mean =81/9 =9 Hence the mean of the terms is 9.

1+3+5+7+9+11+13+15+17=81/9=9

same solution

81/9=9

Sum of Odd numbers = N^2 .
Sum of first 9 odd numbers = 9^2 => 81.
Mean = 81 / 9 = 9

By formula => N^2
Mean of N => N^2 / N => N

First 9 positive odd integers are 1,3,5,7,9,11,13,15 and 17. Mean= $\frac {Sum of all observations}{Total number of obsevations}$ = $\frac {81}{9}$ =9

The first 9 positive odd integers are 1, 3, 5, 7, 9, 11, 13, 15, and 17. When you add them all together you get 81. The question is asking for the mean, so you divide 81 by 9. The final result is 9.

1+3+5+7+9+11+13+15+17=81 81/9=9

(1+3+5+7+9+11+13+15+17)/9=9

The first odd integers are 1,3,5,7,9,11,13,15,17the sum is 81 devided by 9 we find 9

1+3+5+7+9+11+13+15+17=81 81/9=9

The answer is 9. 1+3+5+7+9+11+13+15+17=81 81/9=9

(1+3+5+7+9+11+13+15+17)/9=9

We can approach this problem from several 'paths'. The 'dividing-by-9' part is easy, but the same can't be said for the sum.

A neat way to get this sum is to recognize that the odd numbers form an A.P (Arithmetic Progression) .We see that 1, 3, 5 and so on, have a difference, d, of 2 and start a 1[i.e=a]. There are 9 numbers.

Therefore, sum = n(2a+(n-1)d)(1/2) (We get, putting a=1 and d=2) Sum = 81.

Therefore, mean = sum/n = 81/9 = 9