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*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

Arithmetic Sequence

Series and Sequence

Algebra

But try the problem first...

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

Source

Suggested Reading

American Mathematics Competition

Challenges and Thrills - Pre - College Mathematics

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 )

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