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.

Re: Two series/sequence problems

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

$\displaystyle L=3-\frac{L}{3}$

Solve for $\displaystyle L$.

Re: Two series/sequence problems

Quote:

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

You have a mistake in the second equation.

It should be

$\displaystyle \frac{8(2a + 7d)}{2}=139+115=254$

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.

Re: Two series/sequence problems

Quote:

Originally Posted by

**Plato** You have a mistake in the second equation.

It should be

$\displaystyle \frac{8(2a + 7d)}{2}=139+115=254$

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.

Ah thanks, so from that I have:

16d = -24

d = -3/2

Therefore, [A] 8a -18 = 278

a = 296/8 = 37

Re: Two series/sequence problems

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?

Re: Two series/sequence problems

Quote:

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?

Can you solve $\displaystyle 37-1.5(n-1)<0~?$