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Geometric sequence has U₃= 1458 and U₆= -432 find r..

I’m asked a few questions based on the above initial question. This a four-part question and I believe I have all the correct answers but I believe their solution to be incorrect. I'd like someone just to confirm with me first before I go ahead and report to my tutor and sound like a tit lol.

See attached: their questions here.
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My solution:

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Also, I said due to the nature of the +,- pattern it oscillates but they said because r is negative which also seems wrong to me.

Their solution for c and d are using the wrong r.

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Bear with me uploading the photos
(edited 1 month ago)
Reply 1
Yeah, theyve made a (fairly obvious) mistake.

The sum to infinity with r=2/3 must be greater than the initial value, whereas the sum to infinity with r=-2/3 must be less than the initial value, just by a simple rough cancelling argument.
(edited 1 month ago)
Reply 2
Original post by mqb2766
Yeah, theyve made a (fairly obvious) mistake.
The sum to infinity with r=2/3 must be greater than the initial value, whereas the sum to infinity with r=-2/3 must be less than the initial value, just by a simple rough cancelling argument.

Yeah, a clear rookie mistake. Something I was doing when I first started and something that you’re consistently told to be by careful not to do.

Sorry, what do you mean with the sum to infinity part?
Reply 3
Original post by KingRich
Yeah, a clear rookie mistake. Something I was doing when I first started and something that you’re consistently told to be by careful not to do.
Sorry, what do you mean with the sum to infinity part?

Assuing u1 is positive and r=-2/3 (for example), then
u2 + u3 < 0
u4 + u5 < 0
u6 + u7 < 0
....
So the sum to infinity (and 20 here is close to infinity) must be less than u1.

Conversely if r>0, then all the terms are positive and the sum to infinity is greater than u1.
(edited 1 month ago)
Reply 4
Original post by mqb2766
Assuing u1 is positive and r=-2/3 (for example), then
u2 + u3 < 0
u4 + u5 < 0
u6 + u7 < 0
....
So the sum to infinity (and 20 here is close to infinity) must be less than u1.
Conversely if r>0, then all the terms are positive and the sum to infinity is greater than u1.

I’ll see if I can make sense of what you’re explaining.

In the case r<0, in this case r=-2/3…

U₁ being positive, in this case 3280.5..

Every two consecutive terms added together < 0

Does the r<0 explain the oscillating reasoning?
The sum to infinity will be less than the first term?
Reply 5
Original post by KingRich
I’ll see if I can make sense of what you’re explaining.
In the case r<0, in this case r=-2/3…
U₁ being positive, in this case 3280.5..
Every two consecutive terms added together < 0
Does the r<0 explain the oscillating reasoning?
The sum to infinity will be less than the first term?

You can just write the first few tems out (approximately) and spot the patterns os
3000,-2000,1300,-1000,...
If you pair up u1&u2 and u3&u4 and ... the sum is obviously > 0 as the pair sums are all >0
if you pair up u2&u3 and u4&u5 and u6&u7 and ... the sum is obviously < u1 as the pair sums are all < 0
Its hardly rocket science as the infinite sum formula is a/(1-r) so if 0<r<1 then (1-r)<1 and the sum is > a as youre dividing by a term < 1. If -1<r<0, then 1-r>1 and the sum is < a as youre dividing by a term > 1.

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