Question

the first four terms in an arithmetic sequence are $x y$, $x - y$, $xy$, and $x/y$, in that order. what is the fifth term?

Answer 1

When the first four terms in an arithmetic sequence are x+y, x-y, xy and x÷y, then the fifth term will be $\frac{123}{40}$

A sequence of numbers in which every term (except the first term) is obtained by adding a constant number to the previous term is called an arithmetic sequence

In given arithmetic sequence

The first term = x+y

The second term = x-y

Then, the common difference= (x-y)-(x+y)

x-y-x-y= -2y

The third term = x-y+(-2y)

x-y-2y= x-3y

In question third term is given as xy

So both are equal

x-3y=xy

-3y=xy-x

-3y=x(y-1)

x= $\frac{-3y}{y-1}$

Similarly fourth term will be

x-3y-2y=x-5y

$x-5y=\frac{x}{y}$

substitute the value of x  in the equation

$\frac{-3y}{y-1}-5y=\frac{\frac{-3y}{y-1} }{y} \\\frac{-3y}{y-1} -5y=\frac{-3}{y-1}\\ -3y-5y(y-1)=-3\\-3y-5y^{2}+5y=-3\\ -5y^{2} +2y+3=0\\5y^{2} -2y-3=0$

Split the middle term and factorize it

$5y^{2}+3y-5y-3=0\\y(5y+3)-1(5y+3)=0$

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

$y=1,\frac{-3}{5}$

y cannot be equal to 1, because the third term and fourth term will become x.

$y=\frac{-3}{5}$

substitute the value of y in the equation of x
$x=\frac{-3y}{y-1} \\=\frac{-3\frac{-3}{5} }{\frac{-3}{5}-1 }\\ =\frac{-9}{8}$

The fifth term= $\frac{x}{y}+(-2y)=\frac{x}{y} -2y$

$\frac{\frac{-9}{8} }{\frac{-3}{5} }-2\frac{-3}{5} =\frac{123}{40}$

Hence, when the first four terms in an arithmetic sequence are x+y, x-y, xy and x÷y, then the fifth term will be $\frac{123}{40}$

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