## Problem – Arithmetic Sequence (AMC 10)

*The first four terms in an arithmetic sequence are \( x+y\) , \( x-y\) xy , \(\frac {x}{y}\) in that order. What is the fifth term?*

- \(\frac{12}{110}\)
- \(\frac {123}{40}\)
- \(\frac {16}{17}\)
- 20

**Key Concepts**

Arithmetic Sequence

Series and Sequence

Algebra

## Check the Answer

But try the problem first…

Answer: \(\frac {123}{40}\)

American Mathematics Competition

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First hint

Here is the first hint to start this sum:

There is a very easy method to do this sum

At first we can try to find the difference between two consecutive terms which is

\( (x-y) – (x+y) = -2y \)

So after that we can understand the third and forth terms in terms of x and y.

They can be : \( ( x-3y ) \) and \( ( x – 5y )\)

Now try to do rest of the sum………………………………..

Second Hint

If you got stuck after the first hint you can use this :

Though we from our solution we find the other two terms to be \((x-3y)\) and \((x-5y)\)

but from the question we find that the other two terms are \(xy\) and \(\frac {x}{y}\)

So both are equal.Thus ,

\(xy = x – 3y\)

\( xy – x = – 3y \)

\( x (y – 1) = -3y \)

\( x = \frac {-3y}{ y – 1} \) ……………………………(1)

Again , similarly

\(\frac {x}{y} = x -5y \)

Now considering the equation (1) we can take the value of \(\frac {x}{y}\)

\(\frac {-3}{y – 1}= \frac {-3y}{y -1} – 5y \) …………………….(2)

\( -3 = -3y – 5y(y-1) \)

\( 0 = 5y^2 – 2y – 3\)

\( 0 = ( 5y +3)(y-1)\)

\( y = – \frac {3}{5} , 1\)

We are almost there with the answer. Try to find the answer…..

Final Step

Now from the last hint we find the value of \( y = – \frac {3}{5} , 1\)

But we cannot consider the value of y to be 1 as the 1st and 2 nd terms would be \(x+1\) and \(x-1)\) but last two terms will be equal to x .

So the value of y be \(- \frac {3}{5}\) and substituting the value of y in either \(eq^n\) (1) or \(eq^n\) (2) we get x = -\(\frac {9}{8}\)

so , \(\frac {x}{y} -2y = \frac {9.5}{8.3} + \frac {6}{5} \)

= \(\frac {123}{40} \) (Answer )