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AP and GP! Help! (1 Viewer)

GeWoft

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I am huge fail.

"The numbers 2, a, b are in arithmetic progression and a, b, 9 are in geometric progression. Find a and b."

"For what range of values of y will the series y + y^2 + y^3 +... have a limiting sum? Find this sum if y = 3/4."

Help! What do i do? :(
 

undalay

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GeWoft said:
"The numbers 2, a, b are in arithmetic progression and a, b, 9 are in geometric progression. Find a and b."
b-a = 2-a (From AP)
b = 2a -2 (1)
9/b = b/a (From GP)
9a = b^2 (2)
Sub (1) into (2)

9a = (2a-2)^2
9a = 4a^2 - 8a + 4
0 = 4a^2- 17a + 4
0 = (4a-1)(a-4)
a= 1/4, 4
put in (1)
b=2a-2
b= 6, when a =4
b = -3/2, when a = 1/4




GeWoft said:
"For what range of values of y will the series y + y^2 + y^3 +... have a limiting sum? Find this sum if y = 3/4."
Limiting sum is when |r| < 1.
The r in this is y.
So y must be -1<y<1

Plug into a/1-r.
3/4 / 1 - 3/4
= 3
 

GeWoft

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Thank you so much :)
I was struggling with those 2 for hours, now my assignment is finished!
 

YannY

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GeWoft said:
I am huge fail.

"The numbers 2, a, b are in arithmetic progression and a, b, 9 are in geometric progression. Find a and b."

"For what range of values of y will the series y + y^2 + y^3 +... have a limiting sum? Find this sum if y = 3/4."

Help! What do i do? :(
First question AP: a=2+d (1), b=a+d (2)
(1)-(2): a-b=2-a
2a=2+b (3)
For question GP: a=b/r (4), b=9/r (5). Rearrange EqN (5) making r the subject i.e r=9/b (5).

Substitute (6) in (4): a=b/(9/b)
a=(b^2)/9 (7)

Now subsitute (7) into 3. i.e 2[(b^2)/9]=2+b Times both sides by 9
2b^2-9b-18=0 This is now a quadratic.
b=-1.5 and 6

Substituting these back into equation (3)
When b=-1.5 a=0.25
b=6 a=4
 

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