Quick C2 Question. Watch

Muckamuck
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A geometric series has first term 4 and common ratio r. The sum of the first three terms of the series is 7.

Show that 4r^2 + 4r - 3 = 0.

My working:

7 = \frac{4(r^3 - 1)}{r - 1}



7r - 7 = 4r^3 - 4



4^3 - 7r + 3 = 0.

It appears to be the same equation as the one they asked for but written differently. I don't know how they expect you to arrive at 4r^2 + 4r - 3 = 0.

Can anyone see where I've gone wrong? Thanks.
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Small123
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(Original post by Muckamuck)
A geometric series has first term 4 and common ratio r. The sum of the first three terms of the series is 7.

Show that 4r^2 + 4r - 3 = 0.

My working:

7 = \frac{4(r^3 - 1)}{r - 1}



7r - 7 = 4r^3 - 4



4^3 - 7r + 3 = 0.

It appears to be the same equation as the one they asked for but written differently. I don't know how they expect you to arrive at 4r^2 + 4r - 3 = 0.

Can anyone see where I've gone wrong? Thanks.
7 = \frac{4(r^3 - 1)}{r - 1} \dfrac{r^3-1}{r-1} share a common root
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Oh I Really Don't Care
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If the first term is 4 then the second term is 4r and the third term is 4r^2.

Do you know what they sum to?
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Adam92
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(Original post by Muckamuck)
A geometric series has first term 4 and common ratio r. The sum of the first three terms of the series is 7.

Show that 4r^2 + 4r - 3 = 0.

My working:

7 = \frac{4(r^3 - 1)}{r - 1}



7r - 7 = 4r^3 - 4



4^3 - 7r + 3 = 0.

It appears to be the same equation as the one they asked for but written differently. I don't know how they expect you to arrive at 4r^2 + 4r - 3 = 0.

Can anyone see where I've gone wrong? Thanks.
You don't need to do any of the formula stuff

You know (or should know) that a(r^0) is the first term, a(r^1) is the second and a(r^2) is the third, where a = 4.. so add them together and make it equal 7
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Muckamuck
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(Original post by Small123)
7 = \frac{4(r^3 - 1)}{r - 1} \dfrac{r^3-1}{r-1} share a common root
Thanks.

I like your sig.
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