If x, 2x+2, 3x+3... form a geometric sequence, what are the numerical values of the 1st six terms?

Question

This problem has me banging my head...and I have a math minor...I am trying to help my brother solve it for homework.

So far I have tried ratios, a sub n, the sum of the geometric sequence...no avail.

I found something online that shows the fourth term of this is equal to -27/2. This did not get me anywhere.

For Puggy...I went that route...and got an imaginary number. Same by going the route of solving ratios...x^2+x+4=0, this yields an imaginary term.

I am truly kicking myself now...I took everything to the right to leave things positive...I should have gone the other way. Thank you so much, Puggy!

Answer 1

Remember that in a geometric sequence, there is a ratio r, and if the first term is a, then the sequence would go
a, ar, ar^2, ar^3, etc....

All we have to know is that the quotient of ANY term with its previous term is going to be the same throughout the entire sequence.

This means that

[the third term] / [the second term] = [the second term] / [the first term]

(3x + 3) / (2x + 2) = (2x + 2) / (x)

Multiply both sides by x(2x + 2) to get rid of all fractions, to get

(3x + 3)x = (2x + 2) (2x + 2)

Now, expand both sides.

3x^2 + 3x = 4x^2 + 8x + 4

Now, bring everything to the left hand side.

-x^2 - 5x - 4 = 0

Multiply both sides by -1, to get

x^2 + 5x + 4 = 0

Now, factor this.

(x + 4) (x + 1) = 0
So we get two solutions: x = {-4, -1}

Although we have x, we don't have r. We now solve for r by taking the quotient of (2x + 2) / x

In the case of x = -4, r = (2(-4) + 2)/(-4) = (-8 + 2)/(-4) = 3/2
In the case of x = -1, r = (2(-1) + 2)/(-1) = (-2 + 2)/(-1) = 0

To get the first 6 terms, all you have to do is start with your first term (which is either going to be -4 or -1), and then keep multiplying the ratio.

You get these two sequences:
1) -4, -6, -9, -27/2, -81/4, -243/8 {for x = -4, r = 3/2}
2) -1, 0, 0, 0, 0, 0

However, we must discard the second sequence because we can't have a geometric sequence with r = 0.

Therefore, the only sequence is

1) -4, -6, -9, -27/2, -81/4, -243/8

Answer 2

Geometric sequence means that the third number is y times the second number and that the second number is y times the first number.

So:

xy = 2x + 2 [equation 1]
(2x + 2)y = 3x + 3 [equation 2]

So now you have two sets of equations to help you solve out y.

Let's simplify the second equation first:
2xy + 2y = 3x + 3. Let's leave it like this for now and call it [equation 3]

Substitute the first equation into the equation 3 above:
2(2x + 2) + 2y = 3x + 3
4x + 4 + 2y = 3x + 3
x + 1 + 2y = 0
x = -1 - 2y [equation 4]

Substitute equation 4 into equation 1 and solve:
(-1 - 2y)y = 2(-1 - 2y) + 2
-y - 2y²= -2 - 4y + 2
-2y² - y = -4y
-2y² + 3y = 0
-y(2y - 3) = 0
y = 0 or y = 3/2

Now go solve for x for each of the two possible y values:
xy = 2x + 2
(2x + 2)y = 3x + 3

When y = 0 --> x = 0. This doesn't seem valid
When y = 3/2 --> 3x/2 = 2x + 2
0 = x/2 + 2
x = -4

So the first six numbers of the series is:
x, xy, xy², xy³, ..., xy^5 =
First number is x = -4 like we solved for.
Second number is xy = (-4)(3/2)
Third number is xy^2 = (-4)(3/2)(3/2). Etc.
So the first six numbers in this geometric sequence are:
-4, -6, -9, -27/2, -81/4, -243/8

I might have made a mistake in the number crunching part of the equation, so when you help your brother, you might want to double check to make sure everything's all right.

Answer 3

A geometric sequence is expressed in the form
a, ar, ar², ar³, ar^4, etc....

You have a constant a, and a ratio r. For any two terms

ar^n/[ar^(n-1)] = r

the ratio = r.

But what do we have?
Dividing the third term by the second we get:

(3x + 3)/(2x + 2) = 3/2

Dividing the fourth term by the third we get:

(4x + 4)/(3x + 3) = 4/3

More generally, dividing the nth term by the (n-1)th we get:

(nx + n)/((n-1)x + (n-1)) = n/(n-1)

The ratio changes depending on which two consecutive terms you divide. So this is NOT a geometric sequence.

Furthermore, I would point out, that if x is an integer, all of the terms in the sequence are integers. None can be fractions.

Answer 4

Do it in a simple way.

By geometric mean,

(2x+2)^2 = x(3x+3)

which can be simplified to

x^2 + 5x + 4 = (x + 4)(x + 1) = 0

Therefore, x = -4 or -1

Starting with x = -4, we have
-4, -6, -9, -27/2, -81/4, -243/8

Starting with x = -1, we have
-1, 0, 0, 0, 0, 0

Answer 5

case1
if a,b,c are in contionuous g.p
b^2=a*c
using this
we get either x=-1or x=-4
case 2
they are in any order
assume that m, nth term of a g.p is 2x+2, 3x+3...
x is the 1st term
comm ratio=(2/3)^1/(n-m)

x*(2/3)^n/(n-m)=2x+2
x*(2/3)^m/(n-m)=2x+2
solving for x in 1 and eq with 2nd
2^(m+1)/3(^n-m)=2^(n)/3^(n-m+1)
2^(m-n+1)=3^(-1)
lhs is a pure factor of 2(if exponent is an integr)
rhs is a pure factor of 3 if exponent is an integr)
clearly lhs!=rhs if exponent is an integr)
n,m are irrational no
but term cannot be irraational


IMPLIES THERE ARE ONLY 2 SOLN FOR THE PROBLEM X=-1
X=-4




.....

Answer 6

well if x=1 then its 1, 4, 6, 8, 10, 12

if x=2, then its 2, 6, 9, 12, etc

it depends on what you let x equal to. i dont think the question is worded well.

Answer 7

x, 2x + 2 , 3x+3,................ is a geometric sequence
a, ar , ar² , ar³,...................
WE HAVE
a = x
ar = 2x + 2
xr = 2(x+1).....(I)
xr² = 3(x+1)....(II)
if we devide (II) by (I) we get
r = 3/2
so we put r = 3/2 in equation (I)
3/2 x = 2 (x+1)...multiply*2
3x = 4x + 4
-x = 4
x = -4
so a = x = -4
so the sequence wanted is
(-4 , -6 , -9 , -27/2 , -81/4 , -243/8,..........)

Answer 8

Im sorry I dont know.