kryton1212
a and b are +ve integers. a,-4,b form a geometric sequence, and -4,b,a form an arithmetric sequence.
(a) Find the value of ab.
(b) Find the values of a and b.
(c) (i) Find the sum to infinity of the geometric sequence a,-4,b,... .
(ii) Find the sum to infinity of all the terms that are +ve in the geometric sequence a,-4,b.
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kryton1212
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@hartnn
hartnn
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if x,y,z form an geometric sequence,
then \(y^2 =xz\)
so, what about
a,-4 and b ?
kryton1212
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ab=(-4)^2=16
why?
kryton1212
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it is also fixed?
hartnn
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i can get you a simple explanation here...
each next term of geometric sequence is MULTIPLIED by the common ratio 'r'
so, in general, terms are a,ar,ar^2,ar^3.....
when we have just 3 terms, we have a, ar, ar^2
for this (1st term * 3rd term) = (2nd term)^2 (you can check)
kryton1212
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okay, i understand
hartnn
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or in other words, they form a proportion, \(\dfrac{1st \: term}{2nd \: term}=\dfrac{2nd \: term}{3rd\: term}\)
hartnn
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16 is correct :)
now if the terms are in arithmetic sequence,
then 2* 2nd term = 1st term + 3rd term
use this...
kryton1212
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wow, i like the formulae
hartnn
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or if you don't want to remember the formula, you should know that difference between the terms is constant in arithmetic sequence
so, difference between b and -4 = difference between a and b
b - (-4) = a-b
you'll get same equation :)
kryton1212
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yup, got it
hartnn
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tell me what u get for a and b when you get it....2 equations, 2 unknowns...
kryton1212
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-4+d=b
b+d=a
ar=-4
ar^2=b
are the equations correct?
hartnn
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ar = -4 ? how ?
kryton1212
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i don't know, i just guess.....it's written couple of days ago. i forgot how...
hartnn
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forget about the 'd' and 'r' now
we have 2 equations in a and b
ab = 16
and from b-(-4) = a-b, we have 2b =a-4
you got these ? how..
kryton1212
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uh, you said this moment ago. got it.
a=8 , b=2
hartnn
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those values are correct.
so b) is done right ?
now c) part
can you find the common ratio in 8,-4,2,....?
kryton1212
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*-2
hartnn
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b=2 was correct....
kryton1212
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common ratio is -2
hartnn
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ohh! yes, r= -2 :)
first term = a1 = 8
SUm to infinite is just \(\large \dfrac{a_1}{r-1}\)
kryton1212
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what? -8/3?
hartnn
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ohh....the r is not -2 :P
kryton1212
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-1/2 ?
hartnn
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yes.
r = 2nd term / 1st term = 3rd trm/ 2nd term =.... = -4/8 = -1/2
kryton1212
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-16/3 ...
hartnn
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since r is LESS than 1, we use
\(\huge S_\infty = \dfrac{a_1}{1-r}\)
kryton1212
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lol
hartnn
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yes, there are 2 formula, depending on r less than 1 or greater than 1
kryton1212
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i will remember. I just forgot at that moment..
hartnn
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try the last aprt ?
kryton1212
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i hate doing +ve / -ve ....
hartnn
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remove all the negatives...
8,-4,2,-1,1/2 ....
to
8,2,1/2 ....
now whats the common ratio r =... ?
kryton1212
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oh, i know what's the world happening!
r=1/4
hartnn
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yessss!
hartnn
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and since 1/4 is less than 1
you use
a/ (1-r)
kryton1212
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32/3
kryton1212
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million thanks
hartnn
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\(\huge \color{red}\checkmark\)
hartnn
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welcome ^_^