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Geometric series

Hi everyone,

I've got 3 geometric series questions that I'm struggling with. It would be great if anyone could give me a few pointers.

1)A geometric series has has 4th term equal to 8 times the 7th term. First term is 1024. Find the sum of the first 15 terms.

OK, so as I know the first term, I need the common ratio. I started with ar3=8ar6ar^3=8ar^6 but couldn't go on from there as I don't know any of the values except for a=1024.

2)The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positive.

OK, so this is what I did:
5468.752000=r3=2.734...r=2.7343=1.398\frac{5468.75}{2000}=r^3=2.734...\Rightarrow r=\sqrt[3]{2.734}=1.398 however it's not the right answer.

3)The sum of the first n terms of the geometric series is: 3-6+12-... is 129. Find n.

So:
129n=3(1(2n))1(2)129_n = \frac{3(1-(-2^n))}{1-(-2)} but I really don't know how to derive n when I don't know r!

I'm stuck :frown:

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Reply 1
Original post by gavinlee
Hi everyone,

I've got 3 geometric series questions that I'm struggling with. It would be great if anyone could give me a few pointers.

1)A geometric series has has 4th term equal to 8 times the 7th term. First term is 1024. Find the sum of the first 15 terms.

OK, so as I know the first term, I need the common ratio. I started with ar3=8ar6ar^3=8ar^6 but couldn't go on from there as I don't know any of the values except for a=1024.


Question 1,

ar3=8ar6 ar^3=8ar^6

Now divide both sides by 'a' to get, r3=8r6    8r6r3=0 r^3=8r^6 \implies 8r^6-r^3 = 0
Find 'r' from it.

For the sum, just use the formula, a(1rn)1r \displaystyle \frac{a(1-r^n)}{1-r}
For Q1

only using the equation ar3=8ar6a r^3 = 8 a r^6 you should notice that you have 1 equation and 1 unknown, (silent hooray) this is good because this means you are guaranteed a solution.

Dividing through by aa and r3r^3 leaves you with the equation 1=8r31= 8 r^3 which you can then solve to find r=0.5r = 0.5

Now input your known values into the formula, and you should get an answer of Sn=2047.9375S_n = 2047.9375


For Q3

r=2,a=3r = -2, a = 3

hence using Sn=a1rn1rS_n = a\frac{1-r^n}{1-r} we get:

129=31(2)n1(2)129 = 3\frac{1-(-2)^n}{1-(-2)}

129=31(2)n3129 = 3\frac{1-(-2)^n}{3}

129=1(2)n129 = 1-(-2)^n

128=(2)n128 = -(-2)^n

128=(2)nn=7-128 = (-2)^n \Rightarrow n=7



I need to go to uni, so hopefully someone will help with Q2 or maybe this will be enough for you to work out how to do it. Good luck :smile:
Reply 3
Original post by bencrossley1
For Q1

only using the equation ar3=8ar6a r^3 = 8 a r^6 you should notice that you have 1 equation and 1 unknown, (silent hooray) this is good because this means you are guaranteed a solution.

Dividing through by aa and r3r^3 leaves you with the equation 1=8r31= 8 r^3 which you can then solve to find r=0.5r = 0.5

Now input your known values into the formula, and you should get an answer of Sn=2047.9375S_n = 2047.9375


For Q3

r=2,a=3r = -2, a = 3

hence using Sn=a1rn1rS_n = a\frac{1-r^n}{1-r} we get:

129=31(2)n1(2)129 = 3\frac{1-(-2)^n}{1-(-2)}

129=31(2)n3129 = 3\frac{1-(-2)^n}{3}

129=1(2)n129 = 1-(-2)^n

128=(2)n128 = -(-2)^n

128=(2)nn=7-128 = (-2)^n \Rightarrow n=7



I need to go to uni, so hopefully someone will help with Q2 or maybe this will be enough for you to work out how to do it. Good luck :smile:


Don't mind but please avoid giving full solutions. You are not very active member of TSR so might not be aware of the rules, full solutions are considered a last resort.
Reply 4
Thanks Raheem and Ben.


Does anyone know why this is wrong:


5468.752000=r3=2.734...r=2.7343=1.398\frac{5468.75}{2000}=r^3=2.734...\Rightarrow r=\sqrt[3]{2.734}=1.398

For this question:

The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv
Reply 5
Original post by gavinlee
Thanks Raheem and Ben.


Does anyone know why this is wrong:


5468.752000=r3=2.734...r=2.7343=1.398\frac{5468.75}{2000}=r^3=2.734...\Rightarrow r=\sqrt[3]{2.734}=1.398

For this question:

The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv


For this question you have to use the formula,
Sn=a(1rn)1r \displaystyle S_n=\frac{a(1-r^n)}{1-r}

S4=2000(1r4)1r=5468.75    r42.734375r+1.734375=0 \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

I don't know why we are getting such a complex equation.

What is the answer of this question?
Original post by raheem94
For this question you have to use the formula,
Sn=a(1rn)1r \displaystyle S_n=\frac{a(1-r^n)}{1-r}

S4=2000(1r4)1r=5468.75    r42.734375r+1.734375=0 \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

I don't know why we are getting such a complex equation.

What is the answer of this question?


Does seem like a very awkward question.

One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

then (1+r^2)(1+r)=175/64

(16+16r^2)(4+4r)=175

then r = 3/4 works by inspection.

VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.
Reply 7
Original post by hassi94
Does seem like a very awkward question.

One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

then (1+r^2)(1+r)=175/64

(16+16r^2)(4+4r)=175

then r = 3/4 works by inspection.

VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.


I actually have tried the way you are saying, by adding the first 4 terms. But i think that some value might be wrong in the question, or there might be something missing here. Hence, i left it, i don't see any reason for a GP question to be so awkward.

By the way, congrats for winning the member of the month award.
Well I mean its clear that r is greater than 1/2 and smaller than 1 so solving it iteratively probably wouldn't be a massive pain but yes it does seem very odd for a GP question


And thank you! :biggrin:
Reply 9
I'm glad I'm not the only one who thinks it's awkward! I haven't got the answer on me at the moment but when I get back home I'll post it.

Thanks very much for trying to help me.
Gav.
Reply 10
Original post by gavinlee
Thanks Raheem and Ben.


Does anyone know why this is wrong:


5468.752000=r3=2.734...r=2.7343=1.398\frac{5468.75}{2000}=r^3=2.734...\Rightarrow r=\sqrt[3]{2.734}=1.398

For this question:

The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv


The answer to this is 34\frac{3}{4}.
Reply 11
Original post by gavinlee
The answer to this is 34\frac{3}{4}.


a+ ar+ar^2+ar^3=5468.75

2000+ 2000r+ 2000r^2+ 2000r^3= 5468.75

r^3+ r^2+ r- 111/64=0

(cheat here... you know r=3/4 is a factor).
(edited 11 years ago)
Reply 12
I remember this question, C2 on AQA right?
Reply 13
Original post by raheem94
For this question you have to use the formula,
Sn=a(1rn)1r \displaystyle S_n=\frac{a(1-r^n)}{1-r}

S4=2000(1r4)1r=5468.75    r42.734375r+1.734375=0 \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

I don't know why we are getting such a complex equation.

What is the answer of this question?


I get a pretty reasonable answer...
Original post by f1mad
Un= a*r^(n-1)

a= 2000 here


What's your point? :s
Reply 15
Original post by hassi94


One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

then (1+r^2)(1+r)=175/64

(16+16r^2)(4+4r)=175

then r = 3/4 works by inspection.

VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.


No, it factorises as a cubic :smile:.

Subtract 5468.75/2000 from both sides.
Reply 16
Original post by hassi94
What's your point? :s


Ignore that, I realised you'd divided by 2000 :tongue:.
Original post by f1mad
No, it factorises as a cubic :smile:.

Subtract 5468.75/2000 from both sides.


But how does one just know 3/4 is a factor as you said.
Reply 18
Original post by hassi94
But how does one just know 3/4 is a factor as you said.


You can cheat I guess :tongue:.

Or you can find possible factors of 111/64

+-(1,3,37,111)/ +-(1,2,4,8,16,32,64).

Long process, but it works.
Original post by f1mad
You can cheat I guess :tongue:.

Or you can find possible factors of 111/64

+-(1,3,37,111)/ +-(1,2,4,8,16,32,64).

Long process, but it works.


In the end you're still doing what I'm doing, by inspection, which I feel is a bit silly...

Didn't mean silly to do, silly of aqa to require it :tongue:
(edited 11 years ago)

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