A Sequence of Increasing Averages

The first and second numbers have average one, yet is both at least one, since they are positive integers. Therefore they are both equal to 1. Let $a_n$ denote the $n$ th number. The average of $a_2=1$ and $a_3$ is $2$ , and thus $a_3=3$ . We prove that $a_n=n$ if $n$ is odd, and $a_n=n-1$ otherwise. We do so by induction. Suppose we know the values of $a_1,...a_{k-1}$ . If $k$ is even, then by our induction assumption we know $a_{k-1}=k-1$ . It follows that $k-1+a_k=2(k-1)$ since their average is $k-1$ . Hence $a_k=k-1$ . In the other case, $k$ is odd. Then $a_{k-1}=k-2$ , and so $k-2+a_k=2(k-1)$ again, which implies $a_k=k$ . It follows by induction that our assumption is true, and we now see that $a_{100}=99$ , which is our answer.

Let the integers be $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8\dots$ , where $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8\dots \in Z^+$

Then $\frac{a_1+a_2}{2} = 1$

Keeping in mind the condition that $a_1,a_2 \in Z^+$ , the only possible solution is for $a_1,a_2 = 1$

Now the average of $a_2$ and $a_3$ is $2$ . $\Rightarrow \frac{a_2+a_3}{2} = 2$ $\Rightarrow a_2+a_3 = 4$ $\Rightarrow a_3 = 4 - 1 = 3$

Again, the average of (a_3) and $a_4$ is $3$ . $\Rightarrow \frac{a_3+a_4}{2} = 3$ $\Rightarrow a_3+a_4 = 6$ $\Rightarrow a_4 = 6 - 3 = 3$

Similarly, we can prove the series to be $1,1,3,3,5,5,7,7,9,9 \dots$

Terms $1$ and $2$ are $1$

Terms $3$ and $4$ are $3$

Terms $5$ and $6$ are $5$

The above can be generalized for the given series.

$a_n = \left\{ \begin{array}{l l} n - 1 & \quad \text{if \$n\$ is even}\\ n & \quad \text{if \$n\$ is odd}\\ \end{array} \right.$

Now substituting $n = 100$ , we get $a_n = 99$

Let x 1, x 2, x 3, ………x 100 be the 100 positive integers written in a row. Then , Average of the first and second number is 1 implies (x 1 + x 2)/2=1 i.e., x 1 + x 2 = 2 or x 1 = x 2= 1 Similarly, average of second and third number is 2 implies (x 2 + x 3)/2=2 i.e., x 2 + x 3= 4 or x 3= 3 Similarly , average of third and fourth number is 3 implies (x 3+x 4)/2=3 i.e., x 3+x 4=6 or x 4=3 Average of fourth and fifth number is 4 implies (x 4+x 5)/2=4 i.e., x 4+x 5=8 or x 5=5 Average of fifth and sixth number is 5 implies (x 5+x 6)/2=5 i.e., x 5+x 6=10 or x 6=5 So the row looks like 1,1,3,3,5,5,……. i.e., odd numbers appearing twice hence the 100th number has to be the 99.

Let the sequence is a 1,a 2,a 3,....a 99,a 100. The terms of the sequence are positive integers. So the only pattern for the first condition, i.e (a 1+a 2)/2 = 2 is (1+1)/2=2. Thus a 1=1, a 2=1, then the second condition {(a 2+a 3)/2=2} only fulfiled by (1+3)/2=2, then the second condition {(a 3+a 4)/2=3} only fulfiled by (3+3)/2=3, then the second condition {(a 4+a 5)/2=4} only fulfiled by (3+5)/2=4, then the second condition {(a 5+a 6)/2=5} only fulfiled by (5+5)/2=5, then and so on.. We get the sequence is 1,1,3,3,5,5,7,7,.... Thus a 1,a 3,a 5,a 7,...=1,3,5,7,... and also a 2,a 4,a 6,...=1,3,5,7,... We define a mapping from {a 2,a 4,a 6,...} to {1,3,5,...} by a 2n ---> 2n-1. Then a 100=a (2x50) then mapped to 2(50)-1=99.

By finding the first two number, it is easy to find the 100th number. From the first equation (a+b)/2=1, a+b=2. Because the numbers all have to be positive integers, a and b must be 1 or bigger. However, because the addition of a and b is 2. The only possibility can be a=b=1. Then find the pattern which keeps going 1,1,3,3,5,5,7,7, ...... Which means that the odd numbers repeat twice every time. So 99th and 100th number would be 99.

If average of two positive integers are 1 then the numbers can not be anything except 1 and 1. (As 0 or negative integers are not allowed)

Now the average of 1 and X is 2 then X has to be 3. Now the average of 3 and Y is 3 then Y has to be 3. Now the average of 3 and Z is 4 then Y has to be 5.

Then the sequence would be 1,1,3,3,5 ......

SO. nth term= n if n is odd and n-1 if n is even.

So, 100th term= 100-1 =99

Let [n] be the nth number in the sequence. [1]+[2]=2, [1]=[2]=1 Since [n]+[n+1]=2n and [n+1]+[n+2]=2n+2, [n+2]-[n]=2, [n+2]=[n]+2 where 1<=n<=98 and n is an integer. [100]=[98]+2=[96]+2(2)=...=[2]+49(2)=1+49(2)=99

Let the first integer be A1, the second be A2, the third be A3 and so on. (A1+A2):2 = 1 so A1+A2 = 2, A2+A3 = 4, A3+A4 = 6, and so on. Look at A1+A2 = 2. A1 and A2 must be positive integer. So, A1 must be 1 and A2 must be 1 too. A2 + A3 = 4. So, A3 = 3. A4 = 3 too. So, A100 = A99 = 99

n {1} is the average of i {1} and i {2} because 0 is not a positive integer the only possible solution is i {1} and i {2} are both 1. n {2} is 2 and the average of i {2} and i {3} so to find i {3} you put the known values into an average of the two integers 2= \frac{1+i {3}}/{2} solving this i {3} = 3. This repeats with the next two integers being 3. So the list of integers is 1,1,3,3,5,5,7,7,... every odd number twice in a row. The integers corresponding to every odd number are itself and the integers corresponding the every even number and the number + and - n. i {99} + i {100} = 99 because this 99 is odd we know that both integers are 99 so i_{100} = \boxed{99}

Average of 1 and 1 is 1 and no other number satisfies it. Average of 1 and 3 is 2. Average of 3 and 3 is 3. Average of 3 and 5 is 4 and so on.

You will see that the pattern is $1,1,3,3,5,5,...........$ .

There are 50 distinct numbers in the series and the last one is 99 which can be verified from the last statement which says $the$ $average$ $of$ $the$ $99th$ $and$ $100th$ $number$ $is$ $99$ .

My logic for this question: (a trial and error method)

If n is any odd positive integer,

n = $\frac{n+n}{2}$

n+1 = $\frac{n+n}{2}$ +1 ...................[adding 1 to both the sides]

n+1 = $\frac{n+n}{2}$ + $\frac{2}{2}$

n+1 = $\frac{n+n+2}{2}$

n+1 = $\frac{(n)+(n+2)}{2}$

Taking (n=1)

1 = $\frac{1+1}{2}$

2 = $\frac{(1)+(1+2)}{2}$

Similarly,

99 = $\frac{99+99}{2}$

Therefore, the sequence can be written as: n, n, n+2, n+2, n+4, n+4, ....

Hence, the given sequence is: 1,1,3,3,5,5,.................,99,99.

Let nth number be x n... Therefore, x 100 = 99X2 - x_99

x 100 = 99X2 - ( 98X2 - ( 97X2 - ( 96X2 - ( 95X2 .........- ( 1X2 - x 1 ))))))

x 100 = 2 ( 99 - 98 + 97 - ...... - 2 + 1) - x 1

x 100 = 100 - x 1

Concluding from this, the series can be like -

1, 1, 3, 3, 5, 5, 7, ................ 99, 99

x_1 = 1(smallest +ve integer)

therefore, x_100 = 99

Given: 1)x+y/2=1 so, x=y=1 2) y+z/2=2 from above, 1+z=4 i.e z=3

3) Z+a/2=3
from (2),3+a=6 i.e, a=3

Here, where the average is an odd number, equal numbers are added i,e x+x/2. since, the average 99 is an odd number, 99th and 100th number must be same and hence the 100th number is 99.

The Average of 1st and 2nd number is = 1

The Total of 1st and 2nd number = $1 \times 2 = 2$

So the number may be $\boxed{0,2} or \boxed{1,1}$

The Average of 2nd and 3rd number is = 2

The Total of 2nd and 3rd number is = $2 \times 2 = 4$

So, the number may be $\boxed{0,4}; \boxed{1,3} or \boxed{2,2}$

The Average of 3rd and 4th number is = 3

The Total of 3rd and 4th number is $3 \times 2 = 6$

So, The number may be $\boxed{0,6}; \boxed{1,5}; \boxed{2,4} or \boxed{3,3}$

*From that we can get the 1st number = 1; 2nd number = 1; 3rd number = 3; 4th number = 3 *

So, The sequence is $1, 1, 3, 3, 5, 5........ 99, 99$

And the Average of 99th and 100th number is = 99

So, The 100th number is $\boxed{99}$

I couldn't come up with a mathematical solution. So I gave up and made me computer do it for me. So this is for those who like programming. The series is printed out on console and the last number is the 100th term. The answer if 99

// include iostream statement goes here

//coded in c++

using namespace std;

int main () {

int counter (0);

for (int i = 1; i <= 100; i++) {

    if (counter == 100) { break; }

    if (i % 2 != 0) {

    cout << i << ", " << i << ", ";

    counter += 2;

    }

}

return (0);

}

if we look closely it seems that sequence of odd no. are same,
and sequence of even no. are less 1
so 100th term is 100-1 = 99

Sol:since we know that average =sum of number/total number&given that total number=2 ,average =99 so 100th number=99

n1n2n3na....n100

n1n2 =1 n3n4= 3 n5n6=5 hence its according to first n. Therefore n99n100 = 99

let a=first number,,b=2nd number, c=99th number and d=100th number we know that (a+b)/2=2,thus a+b=2.......since a and b must be positive integers thus a=1 and b=1 also.then the last two number must be also same number. (c+d)/2=99 c+d=198 since c=d, 198/2=99

99

The average of the second and third numbers is 2. and the formula to find the average of 2 number is (first number + Second number)/2. Its means the first number and second number is same. Similarly the 99th and 100th number is same that is 99.

The number series is 1, 1, 3, 3, 5, 5, 7, 7,.... Note that all integers that are odd are in the same place as their value: (i.e. 1 is in the first spot, 3 is in the third spot, 5 is in the fifth spot, etc.). That means that the 99th number in the series is 99. Therefore the next one, 100, is 99.

Follow the question to write out the numbers: 1,1,3,3,..... We can deduce that the pattern follows and if the average of the 99th and 100th number is 99, then for every odd n, it would be n. and (n+1) would be n.

(1+1)/2 = 1, (1+3)/2= 2, (3+3)/2=3.......... (99+99)/2=99..