Question

Given a series \( 4,x,y,z\)
The first three terms are geometric series and the last three terms are arithmetic series. Find the value of \( x+y\)

Answer 1

did u provide the value of contstants

Answer 2

what do you mean?

Answer 3

It's not a unique one.

Answer 4

You can just add the \(z\) according to the values of \(x\) and \(y\).

Answer 5

Using the concept of arithmatic and geometric mean u get.


\(\large \color{black}{\begin{align} x^2& =4y\hspace{.33em}\\~\\
2y& =x+z\hspace{.33em}\\~\\

\end{align}}\)

Answer 6

since the first 3 terms are geometric
$$
\Large { \frac{ x} {4 }= \frac{y}{x}

}
$$

since the last three terms are arithmetic
$$
\Large {
y-x = z - y

}
$$

Answer 7

I got stuck after that step :(

Answer 8

$$
\Large {

x^2 = 4y \\
2y = z + x
\\ \therefore \\
x^2 = 2(2y) \\
x^2 = 2( z + x)\\
x^2 -2x - 2z = 0
}
$$

You can use quadratic formula to solve for x in terms of z

Answer 9

I'll try...

Answer 10

I assume the directions is to find x + y in terms of z

Answer 11

i remember this one!

Answer 12

or try to put \(y=1,2,3\cdots \) there should be a pattern .

Answer 13

I get
\(x=1+\sqrt{1+2z}\) or \(x=1-\sqrt{1+2z}\)

so, \(y=\frac{1+z+\sqrt{1+2z}}{2}\) or \(y=\frac{1-z+\sqrt{1+2z}}{2}\)

so, \(x+y\) will be in term of \(z\), but this is multiple choice question,
the options are
-1 ; 0; 3; 7

Answer 14

if its multiple choice its easier to plug in values
sorry i didnt realize

Answer 15

u should clearly state if it was multiple choice or not otherwise it seems ambigous.

Answer 16

did you get the answer @chihiroasleaf

Answer 17

not yet :(

Answer 18

ok so you have an expression for x + y , in terms of z.
the solution is not unique, you can try plugging in values, the easiest value to plug in is
z = 0

Answer 19

if, z= 0, we get x+y = 3, and the sequence will be 4, 2, 1, 0
right?

Answer 20

$$
\Large {

x^2 = 4y \\
2y = z + x
\\ \therefore \\
x^2 = 2(2y) \\
x^2 = 2( z + x)\\
x^2 -2x - 2z = 0 \\
x = 1 + \sqrt{ 1+ 2z} , ~~1 - \sqrt{ 1 + 2z }
\\ \therefore \\
y = \frac{z + x}{2} \\
\\ \therefore \\
y= \frac{z+1+\sqrt{1+2z}}{2} , ~ \frac{z+1-\sqrt{1+2z}}{2}
\\ \therefore
\\x + y\\
= (1 + \sqrt{ 1+ 2z}) + \frac{z+1+\sqrt{1+2z}}{2} \\
= \frac{2}{2}\cdot (1 + \sqrt{ 1+ 2z}) + \frac{z+1+\sqrt{1+2z}}{2} \\
\\
= \frac{2 + 2\sqrt{1 + 2z} + z + 1 + \sqrt{ 1 + 2z} }{2}
\\
x+y = \frac{ 3\sqrt{1 + 2z} + z + 3 }{2}

\\ \text { if z= 0 } \\
x+y = \frac{3 + 3} {2} = 3
}


$$

Answer 21

yes thats correct

Answer 22

also z = 0 is a logical choice in order to avoid getting an irrational square root number

Answer 23

z = 4 also produces a whole number output

Answer 24

thanks a bunch @perl