science_geeks
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Hi

I'm slightly stuck with this question and I'm struggling to see how to begin please can someone help?

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science_geeks
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RDKGames
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Harold
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mqb2766
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(Original post by science_geeks)
RDKGames
one is an arithmetic sequence (add 1000 each year)
one is a geometric sequence (multiply by 1.05 each year)

Can you write down those two position to term formulae?
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science_geeks
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(Original post by mqb2766)
one is an arithmetic sequence (add 1000 each year)
one is a geometric sequence (multiply by 1.05 each year)

Can you write down those two position to term formulae?
So, is the first one:

Un = a + (n-1)d so, 25,000 + (n-1)1000

Geometric sequence:

Un = ar^n-1 so, 22,000*1.05^n-1 ?? Why have you used 1.05 by the way?
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Muttley79
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(Original post by science_geeks)
So, is the first one:

Un = a + (n-1)d so, 25,000 + (n-1)1000

Geometric sequence:

Un = ar^n-1 so, 22,000*1.05^n-1 ?? Why have you used 1.05 by the way?
If something increase by 5% then you have 105/100 more the next year [100 + 5] which is 1.05
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science_geeks
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(Original post by Muttley79)
If something increase by 5% then you have 105/100 more the next year [100 + 5] which is 1.05
Okay, cool! Thank you
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science_geeks
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(Original post by mqb2766)
one is an arithmetic sequence (add 1000 each year)
one is a geometric sequence (multiply by 1.05 each year)

Can you write down those two position to term formulae?
I've got up to part d, but I'm unsure of what to do after this.. Name:  IMG_3827.jpeg
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simon0
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For part d, you want the total Lewis has earned to be greater than sum value T, so you want the total sum to be greater than T.

Your setup should be this:  \displaystyle \underbrace{\frac{22000 (1 - 1.05^{n})}{(1-1.05)}}_{S_{n}} > T.

Can you take it from there?
Last edited by simon0; 9 months ago
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