Arithmetic Progressions

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VeeN22

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The first two terms of a geometric progression are the same as the first two terms of an arithmetic progression. The first term is 8 and is greater than the second term. The sum of the first three terms of the arithmetic progression is 1.125 less than the sum of the first three terms of the geometric progression. Determine the first three terms of each sequence?

What I only know is GP 8+T2+T3 - AP 8+T2+T3=1.125.
I don't know how to get the terms. Please help.
 
AP=8, x1, x2
GP=8, y1, y2

X1=Y1

8+X1+X2+1.125=8+Y1+YySUB]2[/SUB]
(1.12.5+X1+X2)/Y2=Y1

Can you continue?
 
VeeN22 said:
The first two terms of a geometric progression are the same as the first two terms of an arithmetic progression. The first term is 8 and is greater than the second term. The sum of the first three terms of the arithmetic progression is 1.125 less than the sum of the first three terms of the geometric progression. Determine the first three terms of each sequence?

What I only know is GP 8+T2+T3 - AP 8+T2+T3=1.125.
I don't know how to get the terms. Please help.
Click to expand...

The first 3 terms of the AP may be written as:

[MATH]8,8+d,8+2d[/MATH]
And the first 3 terms of the GP may be written as:

[MATH]8,8r,8r^2[/MATH]
Now, given that the second term in both progressions is the same, we have:

[MATH]8+d=8r\implies d=8(r-1)[/MATH]
And the information regarding the sums of the first 3 terms allow us to write:

[MATH]8+8+d+8+2d=8+8r+8r^2-\frac{9}{8}[/MATH]
Or:

[MATH]16+3d=8r+8r^2-\frac{9}{8}[/MATH]
Now, substitute for \(d\):

[MATH]16+3(8(r-1))=8r+8r^2-\frac{9}{8}[/MATH]
What do you get upon solving this quadratic for \(r\)?
 
MarkFL said:
The first 3 terms of the AP may be written as:

[MATH]8,8+d,8+2d[/MATH]
And the first 3 terms of the GP may be written as:

[MATH]8,8r,8r^2[/MATH]
Now, given that the second term in both progressions is the same, we have:

[MATH]8+d=8r\implies d=8(r-1)[/MATH]
And the information regarding the sums of the first 3 terms allow us to write:

[MATH]8+8+d+8+2d=8+8r+8r^2-\frac{9}{8}[/MATH]
Or:

[MATH]16+3d=8r+8r^2-\frac{9}{8}[/MATH]
Now, substitute for \(d\):

[MATH]16+3(8(r-1))=8r+8r^2-\frac{9}{8}[/MATH]
What do you get upon solving this quadratic for \(r\)?
Click to expand...

I calculated as:
8r^2+8r - 23r - 9/8 -16
= 8r^2 - 15r - 17,125

Hence for r I got 2,68 or -0,80
Correct?
 
VeeN22 said:
I calculated as:
8r^2+8r - 23r - 9/8 -16
= 8r^2 - 15r - 17,125

Hence for r I got 2,68 or -0,80
Correct?
Click to expand...

I get different values for \(r\)...let's go back to:

[MATH]16+3(8(r-1))=8r+8r^2-\frac{9}{8}[/MATH]
Distribute:

[MATH]16+24r-24=8r+8r^2-\frac{9}{8}[/MATH]
Collect like terms:

[MATH]16r=8r^2+\frac{55}{8}[/MATH]
Multiply by 8:

[MATH]128r=64r^2+55[/MATH]
Arrange in standard form:

[MATH]64r^2-128r+55=0[/MATH]
Factor:

[MATH](8r-5)(8r-11)=0[/MATH]
Hence:

[MATH]r\in\left\{\frac{5}{8},\frac{11}{8}\right\}[/MATH]
Now, which root should we use, given that the second term is less than the first in both progressions?
 
MarkFL said:
I get different values for \(r\)...let's go back to:

[MATH]16+3(8(r-1))=8r+8r^2-\frac{9}{8}[/MATH]
Distribute:

[MATH]16+24r-24=8r+8r^2-\frac{9}{8}[/MATH]
Collect like terms:

[MATH]16r=8r^2+\frac{55}{8}[/MATH]
Multiply by 8:

[MATH]128r=64r^2+55[/MATH]
Arrange in standard form:

[MATH]64r^2-128r+55=0[/MATH]
Factor:

[MATH](8r-5)(8r-11)=0[/MATH]
Hence:

[MATH]r\in\left\{\frac{5}{8},\frac{11}{8}\right\}[/MATH]
Now, which root should we use, given that the second term is less than the first in both progressions?
Click to expand...

We use 5/8 because when we substitute r into 8r =8(5/8)
, we get 5 which is less than 8 in both progressions. We can't use 11/8 as 11 is greater than 8.
I hope I got it this time, right?
 
MarkFL said:
Yes, I agree. So, what must \(d\) be?
Click to expand...

d = 5 - 8 = - 3

I then got all the three first terms of both progressions and later subtracted the sum of the GP which is 16,125 - the sum of the AP which is 15 and the answer is exactly 1,125.

Thank you for taking me step by step. I understand it now.
 
VeeN22 said:
d = 5 - 8 = - 3

I then got all the three first terms of both progressions and later subtracted the sum of the GP which is 16,125 - the sum of the AP which is 15 and the answer is exactly 1,125.

Thank you for taking me step by step. I understand it now.
Click to expand...

Good job, specially in verifying everything worked. :)
 
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