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Series help (1 Viewer)

cookiez69

What a stupid name, Nat.
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Find a and b if a,b,1 forms a GP, and b,a,10 forms an AP.

Answer is...
GP:
a= 25/4
b= 5/2

AP:
a=4
b=-2

Also

a) Show that if the first, second and fourth terms of an AP form a geometric sequence, then either the sequence is a constant sequence, or the terms are the positive integer multiples of the first term.
b) Show that if the first, second and fifth terms of an AP form a geometric sequence, then either the sequence is a constant sequence, or the terms are the odd positive integer multiples of the first term.
c) Find the common ratio of the GP in which the first, third and fourth terms form an arithmetic sequence. [Hint: r^3 - 2r^2 + 1 = (r-1)(r^2 - r - 1)
d) Find the GP in which each term is one more than the sum of all previous terms.

Sorry for all the questions, but i'm stuck :\
 
Last edited:

cineti970128

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cookiez man i feel so sorry for you,
these questions are so bothering no one will answer to this

but dw i will haha i have a lot of time to waste at night
 

cineti970128

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i will do one question at a time, my cp legs a lot
1.a)
1st 2nd and 4 th arithemetic sequence
therefore they are
a, a+r , a+3r
if these 3 terms are in GP they should have a geometrical mean
hence a+r/a = (a+3r)/(a+r)
since you are doin 3 unit
you should be able to do some algebra here an find that
r = 0 or a
hence it is a constant sequence ie
a, a, a, a, a, .....
a, 2a, 4a, ...
 
Last edited:

cineti970128

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b) really similar to the first one

you use the GP mean method again and using simple algebra you get
r= 0 or 2a
hence the sequence is
1. a, a, a, a, ....
which is the constant sequence

or

2. a, a+r, a+4r
a, 3a, 9a
which is odd positive integer ---3--- multiples of the first term
 

cineti970128

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c)
you have GP
a, ar, ar^2, ar^3, ....
the first, third and fourth are
a, ar^2, ar^3
for them to be AP
ar^2 - a = ar^3- ar^2

r^3 - 2r^2 +1 = 0
using the hint
(r-1)(r^2-r-1) = 0
hence r = 1 or (1+or - (5)^1/2)/2 using quadratic equation
 

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