Dan's Mean Numbers

The mean of the smallest integer is the integer itself. So we know that the smallest number is 2 .

The mean of two smallest integers is 4. We know the smallest integer is 2, so let's assume the 2nd smallest integer to be x. We have $\frac{2+x}{2}$ = 4. Solving, we get the second smallest integer as 6 .

The mean of the three smallest integers is 6. We know the two smallest integers are 2 and 6, so let's assume the 3rd smallest integer to be x. We have $\frac{2+6+x}{3}$ = 6. Solving, we get the third smallest integer as 10 .

The mean of the four smallest integers is 8. We know the three smallest integers are 2, 6 and 10, so let's assume the 4th smallest integer to be x. We have $\frac{2+6+10+x}{4}$ = 8. Solving, we get the fourth smallest integer as 14 .

From these, we observe that the sequence of integers are 2, 6, 10, 14 and so on. So the 10th integer in this case would be 2 + (10 - 1)/2 = $\boxed{38}$ .

Other way to solve is to directly apply the property.

If the sequence a, b, c, d, e, .... has means of 1st smallest integer, 2nd smallest integers, ... forms a sequence (A.P.) with difference D (in our case, 2), then the actual sequence would also be an A.P. with difference 2D (in our case, 4) and the first number in the sequence would be the value corresponding to mean of 1st smallest integer.

the smallest integer is 2.
because mean increase by 2 so the next integer afrer 2 is plus 4
2 , 6 , 10 , 14 , 18 , 22 , 26 , 30 , 34 , 38 so the largest integer is 38

The mean of the smallest number is 2. Therefore the smallest number is two.

The mean of the two smallest numbers is 4. The mean is a calculation of the middle value. We know there is a difference of two between 2 and 4, and in order for four to be the middle value there must be a difference of two between 4 and the next number. Therefore the next number is 6.

The mean of the three smallest numbers is six. the mean average is calculated by dividing the sum of all the terms by the amount of terms. We know the mean is six, and there are three terms, the first two of which sum to 8. This gives us the statement:

$6 = \frac{2 + 6 + x}{3}$

Which can be rearranged to make:

$18 = 8 + x$

Consequently, the third term must be 10.

Notice that the mean average rises by 2 as we add each term - this means we can use the above method repeatedly to calculate the largest term, but this is an inefficient method. Upon examining our sequence, one might note that it starts at two and increases by four each term - giving us the nth term as:

$4n - 2$

Using this we could easily find the tenth term by doing:

$(4 \times 10) - 2 = \boxed{38}$

Although I was a bit lazy and just added 4 to two ten times :P

The mean of one smallest integer is 2, i.e. it is the smallest integer itself because there exists nothing called 'mean of a single no.'. So, first integer is 2.

Now, mean of first and second integer, i.e. (2+a)/2 = 4 & on solving we obtain second integer (a) as 6. Similarly, (2+6+b)/3 = 8, i.e. on solving b = 10.

Now, if we look at first three integers carefully we can clearly identify a series with a common difference of 4. So, we can list down our 10-integers as: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38.

38 is the answer.

The given problem is in Arithmetic Progression -->so first number(a)=2:
-->common difference(d)=4: -->total numbers{n)=10: --> last term=a+(n-1)d==38

As first number is 2 sum of first and 2nd should be 8(4 times 2) so 2nd number is 6. sum of 2nd and 3rd should be 18( 6 times 3) so 3rd number is 10 and so on we get an A.P. with common difference 4
2 6 10 14 18 22 26 30 34 38