Here,

\(A1\) and \(A3\)are positive integer.

\(A3 = \frac{{A1 + A2}}{2}\)

For A3 to be positive integer, there can be 2 options

1. Both \(A1\)and \(A2\)are odd integer.

2. Both \(A1\) and \(A2\) are even integer.

and \(A3\)= always an even integer.

Let check for \(A4\) & \(A5\)

\(A4 = \frac{{A3 + A2}}{2}\)

and

\(A5 = \frac{{A4 + A3}}{2} = \frac{{3A3 + A2}}{4} ( since, A4 = \frac{{A3 + A2}}{2})\)

Clearly, A5 cannot have a denominator greater than 4 ,

Hence option D

*** try putting values for A3 and A2 ( A3 =even integer and A2 =even or odd integer) , then A5 = all of the options except option D

\(A3 = \frac{A1+A2}{2}\)
\(A4 = \frac{A3+A2}{2}\)
\(A5 = \frac{A4+A3}{2}\)

\(A2 = 2A3 - A1\) (from 1st equation)
\(A5 = \frac{1}{2}(\frac{A3+A2}{2}+\frac{A3}{2})\)
\(A5 = \frac{1}{2}(\frac{3}{2}A3+\frac{1}{2}A2)\)
\(A5 = \frac{1}{2}(\frac{3}{2}A3+\frac{1}{2}(2A3 - A1))\)

Simplifying,
\(A5 = \frac{1}{4}(5A3 - A1)\)
Since, A1 and A3 are positive integers, the expression inside the bracket will always be an integer value. It can be equal to -9, 0, 9, 82 so that options A,B,D and E are possible.
Only option not possible is (D).

In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If \(A1\) and \(A3\) are positive integers, which of the following is not a possible value of \(A5\)?


A. \(\frac{-9}{4}\)

B. zero

C. \(\frac{9}{4}\)

D. \(\frac{75}{8}\)

E. \(\frac{41}{2}\)[/quote]


Here,

\(A1\) and \(A3\)are positive integer.

\(A3 = \frac{{A1 + A2}}{2}\)

For A3 to be positive integer, there can be 2 options

1. Both \(A1\)and \(A2\)are odd integer.

2. Both \(A1\) and \(A2\) are even integer.

and \(A3\)= always an even integer.

Let check for \(A4\) & \(A5\)

\(A4 = \frac{{A3 + A2}}{2}\)

and

\(A5 = \frac{{A4 + A3}}{2} = \frac{{3A3 + A2}}{4} ( since, A4 = \frac{{A3 + A2}}{2})\)

Clearly, A5 cannot have a denominator greater than 4 ,

Hence option D

*** try putting values for A3 and A2 ( A3 =even integer and A2 =even or odd integer) , then A5 = all of the options except option D[/quote]




This problem is ill defined, because the question does not say if A2 is an integer or not. If A2 can be a fractional, then the conclusion in the above argument is wrong:
"Clearly, A5 cannot have a denominator greater than 4"

But the question mentions that A1 and A3 are positive integers.
\(A3 = \frac{(A1 + A2)}{2}\)
If A2 were to be fractional, A3 can't be an integer as it would go against what the question states.
So, A2 can't be fractional.

I looked through the answers and got a bit confused.

Since each term is the average of the previous two terms,


\(A_3=\frac{A_1+A_2}{2} \Longrightarrow A_2=2 y-x\)


Then


\(A_4=\frac{A_2+A_3}{2}=\frac{(2 y-x)+y}{2}=\frac{3 y-x}{2}\)


and

\(
A_5=\frac{A_3+A_4}{2}=\frac{y+\frac{3 y-x}{2}}{2}=\frac{5 y-x}{4}
\)

So


\(A_5=\frac{5 A_3-A_1}{4}\)


Check the options:


\(-\frac{9}{4}=\frac{-9}{4}\) possible

\(0=\frac{0}{4}\) possible

\(\frac{9}{4}\) possible

\(\frac{75}{8}=\frac{37.5}{4}\), not an integer multiple of \(\frac{1}{4}\)

\(\frac{41}{2}=\frac{82}{4}\) possible


Thus, the value that is not possible is:

\( \frac{75}{8} \)