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=8ar6 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: 20005468.75=r3=2.734...⇒r=32.734=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=1−(−2)3(1−(−2n)) but I really don't know how to derive n when I don't know r!
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=8ar6 but couldn't go on from there as I don't know any of the values except for a=1024.
Question 1,
ar3=8ar6
Now divide both sides by 'a' to get, r3=8r6⟹8r6−r3=0 Find 'r' from it.
only using the equation ar3=8ar6 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 a and r3 leaves you with the equation 1=8r3 which you can then solve to find r=0.5
Now input your known values into the formula, and you should get an answer of Sn=2047.9375
For Q3
r=−2,a=3
hence using Sn=a1−r1−rn we get:
129=31−(−2)1−(−2)n
129=331−(−2)n
129=1−(−2)n
128=−(−2)n
−128=(−2)n⇒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
only using the equation ar3=8ar6 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 a and r3 leaves you with the equation 1=8r3 which you can then solve to find r=0.5
Now input your known values into the formula, and you should get an answer of Sn=2047.9375
For Q3
r=−2,a=3
hence using Sn=a1−r1−rn we get:
129=31−(−2)1−(−2)n
129=331−(−2)n
129=1−(−2)n
128=−(−2)n
−128=(−2)n⇒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
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.
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