Two series/sequence problems

Evening all, two questions here, first off I'm just sceptical with my answer for the first question, in great need of a second opinion in terms of my methods etc. The second question I'm really struggling where to start. Help would be much appreciated!

Firstly: The sum of the first four terms of an arithmetic sequence is 139 and the sum of the next four terms is 115. Find a and d:

[A] 4/2(2a + 3d) = 139
4a + 6d = 139

[B] 8/2(2a + 7d) = 115
8a + 28d = 115

Solving simultaneously: 16d = -163
d = -163/16

Substituting into equation [A]: 4a + 6(-163/16)=139
a = 1601/32




Second problem:

Given that, Un+1 = 3 - 1/3Un and U1 = 3

a) Find the values for U2, U3 and U4
b) Find the limiting value of Un as n tends to infinity.

For the second one, let be the limiting value, hence the recursive definition will give you:



Solve for .

Quote:

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Originally Posted by DonGorgon

Firstly: The sum of the first four terms of an arithmetic sequence is 139 and the sum of the next four terms is 115. Find a and d:
[A] 4/2(2a + 3d) = 139
4a + 6d = 139

[B] 8/2(2a + 7d) = 115
8a + 28d = 115

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You have a mistake in the second equation.
It should be


Because it says "sum of the next four terms is 115"
It is the sum of first four plus the sum of the next four terms give the eight.

Quote:

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Originally Posted by Plato

You have a mistake in the second equation.
It should be


Because it says "sum of the next four terms is 115"
It is the sum of first four plus the sum of the next four terms give the eight.

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Ah thanks, so from that I have:

16d = -24
d = -3/2

Therefore, [A] 8a -18 = 278

a = 296/8 = 37

In terms of the series with d = -3/2 and a =37. how would you find the value for n, for which the nth term is the last positive term in the series?

Quote:

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Originally Posted by DonGorgon

In terms of the series with d = -3/2 and a =37. how would you find the value for n, for which the nth term is the last positive term in the series?

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Can you solve