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kmeds16
Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.
Sum of 10 = a * (r^4 - 1) -------- = 4 r- 1
@cwrw238 why is it r^4? I thought it's r^10.
hmm, given is: \[S_n=\frac{1-r^n}{1-r}\] \[S_{10}=4=\frac{1-r^{10}}{1-r}\] \[S_{30-10}=48=\frac{1-r^{20}}{1-r}\]
you have been given the sum upto first 10 terms =4 you have also been given the sum upto first 30 terms = 4 + 48 =52 and you have 2 eqns with 2 variables ->solve for a and r now calculate sum for first 60 terms from that subtract sum of first 30 terms.. this should help..
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a(r^10 - 1) / r - 1 = 4 a(r^30 - 1) / r - 1 = 52 ? @shubhamsrg like this?
solve for a and r.. then find the sum of 60 terms subtract sum of 30 terms from sum of 60 terms
@kmeds16 yep @hitten101 mistake in your formulla in the denominator..
I got, r^10 = 3. this is confusing :/ 10th root of 3?!
my mistake r^10 not r^4
how'd you get that? o.O
@kmeds16 @shubhamsrg yes no exponent in the denominator.. you are right
second equation divided by first equation. hehehehehe
your main aim is not to find r,, your main aim is to find sum.. leave it as r^10 = 3
find the sum of S60 and subtract 52, right?
do this,,this might simplify.. substitue r^10 =3 whereever you can leave r-1 as it is.. you can see a/(r-1) = 4/(r^10 -1) in calculation for sum of 60 terms ,make use of this eqn,, no need to find a.. :)
r^60 we all can find.. hmm.. hope that helped..
solving...hehehe ahm, thanks for the idea..