In the sequence: x+4y ; 2y+2 ;x ;y-1, (consisting of 4 terms) the first three terms form an arithmetric progression, while the last three tern form a geometric sequence. Calculate the values of x and y. Answer(s) can be left in surd form.

Question

In the sequence: x+4y ; 2y+2 ;x ;y-1, (consisting of 4 terms) the first three terms form an arithmetric progression, while the last three tern form a geometric sequence. Calculate the values of x and y. Answer(s) can be left in surd form.

Answer 1

"the first three terms form an arithmetric progression"

so....
2y+2 - (x+4y) = x - (2y+2)
which gives x = 2

"the last three tern form a geometric sequence"
so ...
x/(2y+2) = (y-1)/x , but we know x = 2

2/(2y+2) = (y-1)/2

I get y = ±√3

Answer 2

To find the values of x and y, we will use the given information that the first three terms of the sequence form an arithmetic progression, and the last three terms form a geometric sequence.

Let's start with the arithmetic progression: x + 4y, 2y + 2, x.

In an arithmetic progression, the common difference (d) is equal to the difference between any two consecutive terms. Therefore,

Common difference (d) = (2y + 2) - (x + 4y)

Simplifying this equation gives: d = 2 - x

We also know that the first term (a) is x + 4y, and the second term (b) is 2y + 2.

So, using the formula for the nth term of an arithmetic sequence, we have:

b = a + (n-1)d

Substituting the values, we get: 2y + 2 = (x + 4y) + 2(2 - x)

Simplifying further: 2y + 2 = x + 4y + 4 - 2x

Combining like terms: 3y - x = 2

Now, let's move on to the geometric progression of the last three terms: 2y + 2, x, y - 1.

In a geometric progression, the common ratio (r) is equal to the ratio between any two consecutive terms. Therefore,

Common ratio (r) = (x) / (2y + 2)

Again, using the formula for the nth term of a geometric sequence, we have:

b = a * r^(n-1)

The second term (b) is x, and the first term (a) is 2y + 2.

Substituting the values, we get: x = (2y + 2) * (x / (2y + 2))^(1-1)

Simplifying further: x = (2y + 2) * (x / (2y + 2))^0

Since any non-zero number raised to the power of 0 is 1, we have:

x = (2y + 2) * 1

Simplifying even further: x = 2y + 2

Now we have two equations:

1) 3y - x = 2
2) x = 2y + 2

We can solve this system of equations to find the values of x and y.

From equation 2, we can express x in terms of y:

x = 2y + 2

Substitute this into equation 1:

3y - (2y + 2) = 2

Simplifying the equation:

y - 2 = 2

y = 4

Substitute y = 4 back into equation 2 to find x:

x = 2y + 2

x = 2(4) + 2

x = 8 + 2

x = 10

Therefore, the values of x and y are x = 10 and y = 4.