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# Arithmetic Sequence | AMC 10B, 2003 | Problem No. 24

## 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

Answer: $\frac {123}{40}$

American Mathematics Competition

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

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......................................

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.....

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 )

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