series question help. (1 Viewer)

[bananarama69]

sorry to keep bothering you people, but im in another pickle, silly student teacher doesn't get in my head so here goes:

the numbers 2,a,b are in an arithmetic progression and a,b,9 are in a geometric progression. Find a and b

 

[addikaye03]

sorry to keep bothering you people, but im in another pickle, silly student teacher doesn't get in my head so here goes:

the numbers 2,a,b are in an arithmetic progression and a,b,9 are in a geometric progression. Find a and b

Arithmetic series is T2-T1=T3-T2:

a-2=b-a

2a-b=2 ..(1) or b=2a-2

Geometric series is T2/T1=T3/T2:

b/a=9/b..

b^2=9a...(2)

So (1) into (2), (2a-2)^2=9a

4a^2-8a+4=9a

4a^2-17a+4=0

So Product is 16 and Sum is -17 So factors are -16,-1

4a^2-16a-a+4=0

4a(a-4)-(a-4)=0

(4a-1)(a-4)=0 (therefore a=1/4 and 4)

Sub in for b,c

 

[bananarama69]

im stuck at 4a^2-17a+4 lol

maybe thats not right?

 

[Drongoski]

using quadratic eqn formula, you'll get: a = 4 or 1/4.

 

[bananarama69]

i got a=34 or a=0

 

[bananarama69]

for the quad formula, a=4 b=-17 and c=4 yes?

 

[addikaye03]

for the quad formula, a=4 b=-17 and c=4 yes?

Fix'd my first post to show you the facorisation without quadratic