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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.

Mathematics
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if x,y,z form an geometric sequence, then \(y^2 =xz\) so, what about a,-4 and b ?
ab=(-4)^2=16 why?

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Other answers:

it is also fixed?
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)
okay, i understand
or in other words, they form a proportion, \(\dfrac{1st \: term}{2nd \: term}=\dfrac{2nd \: term}{3rd\: term}\)
16 is correct :) now if the terms are in arithmetic sequence, then 2* 2nd term = 1st term + 3rd term use this...
wow, i like the formulae
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 :)
yup, got it
tell me what u get for a and b when you get it....2 equations, 2 unknowns...
-4+d=b b+d=a ar=-4 ar^2=b are the equations correct?
ar = -4 ? how ?
i don't know, i just guess.....it's written couple of days ago. i forgot how...
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..
uh, you said this moment ago. got it. a=8 , b=2
those values are correct. so b) is done right ? now c) part can you find the common ratio in 8,-4,2,....?
*-2
b=2 was correct....
common ratio is -2
ohh! yes, r= -2 :) first term = a1 = 8 SUm to infinite is just \(\large \dfrac{a_1}{r-1}\)
what? -8/3?
ohh....the r is not -2 :P
-1/2 ?
yes. r = 2nd term / 1st term = 3rd trm/ 2nd term =.... = -4/8 = -1/2
-16/3 ...
since r is LESS than 1, we use \(\huge S_\infty = \dfrac{a_1}{1-r}\)
lol
yes, there are 2 formula, depending on r less than 1 or greater than 1
i will remember. I just forgot at that moment..
try the last aprt ?
i hate doing +ve / -ve ....
remove all the negatives... 8,-4,2,-1,1/2 .... to 8,2,1/2 .... now whats the common ratio r =... ?
oh, i know what's the world happening! r=1/4
yessss!
and since 1/4 is less than 1 you use a/ (1-r)
32/3
million thanks
\(\huge \color{red}\checkmark\)
welcome ^_^

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