Melody

Thankyou Alan  

We will stomp on your army like they are ants.  

I assume that you mean the 20th term is -30

Sn=n/2(a+L)

160=20/2(a+-30)

160=10(a-30)

16=a-30

46=a

a=46

The first term is 46

I have no checked you should though

Tn=a+(n-1)d

-30=46+19d

-76=19d

d=-4

That is good :))

What question?

With what ?

Thanks Alan,

Sorry, I screwed up again  

Hi Sabi,  

There are a couple of ways to do this but I will show you the way that I think is usually easiest.

You want to get rid of the fraction.  If this was an equal sign then you would note that x cannot be 2 and then you would multiply both sides by x-2.

It is a little different with an inequality because you would have to have 2 different scenarios, one for when x-2 is positive and one for when x-2 is negative.   SO

it is much better to multiply both sides by something that you KNOW is positive.

Multiply both sides by (x-2)^2   that HAS to be positive

$$\\\frac{1-x}{x-2}>0\\\\
\frac{1-x}{x-2}\times \frac{(x-2)^2}{1}>0\times \frac{(x-2)^2}{1}\\\\
\frac{1-x}{1}\times \frac{(x-2)}{1}>0\\\\
(1-x)(x-2)>0\\\\\\
--------------\\
$The only way 2 expressions multiply to be positive is if they are both positive or they are both negative.$\\\\
1-x>0\quad and \quad x-2>0 \qquad OR \qquad 1-x<0 \quad and \quad x-2<0\\\\
$ I can continue this if you want but there is a better way.$\\\\
------------------\\\\$$

$$\\(1-x)(x-2)>0\\
$If you let $ y=(1-x)(x-2)\\
$then the statement will be true when $y>0\\
y=(1-x)(x-2)\\
y=-1x^2+3x-2\\
$this is a parabola- the coefficient of $ x^2$ is -1$\\
$So it is concave down.$\\
$The middle bit will be above the x axis where y is positive.$\\
$The roots are x=1 and x=2$\\
SO\\
(1-x)(x-2)>0\quad when\;\;1

This is not as hard as it may at first look.

If you need me to explain anything please ask :)

There isn't one, this is not a GP

if it were a GP then

1/-2 would equal -12/1   and this would  be the common ratio

Yes, sorry Alan I did not realize that our answers were different. 

Alan is totally corect.

This is the formula for the the future value of an ordinary annuity.  This is where the money is put into the account at the END of the time period instead of at the beginning.

It is easy to adjust it for this question.  

Here we have $200 invested at the very beginning so we must ADD 200 plus the interest it will accrue for the whole 30 years  that will be    C(1+i)^n = 200*(1.0025^360)

but NO $200 is  invested at the very end SO we must subtract C = $200 at the end

so we get 

$$\\FV=\left[\frac{(1+i)^n-1}{i}\right]+C(1+i)^n-C\\\\
$If you rearrange this you will find that it is identical to Nauseated's formula.$\\\
FV=$my original answer$+C(1+i)^n-C\\\\
FV=116547.38+200(1.0025)^{360}-200\\\\$$

$${\mathtt{116\,547.38}}{\mathtt{\,\small\textbf+\,}}{\mathtt{200}}{\mathtt{\,\times\,}}{\left({\mathtt{1.002\: \!5}}\right)}^{\left({\mathtt{360}}\right)}{\mathtt{\,-\,}}{\mathtt{200}} = {\mathtt{116\,838.748\: \!442\: \!299\: \!145\: \!509\: \!1}}$$

FV = $ 116838.75  

This is identical to Alan's answer.  And Nauseated also agreed that it is correct.

 ------------------------------------------

Nauseated has gone one more step with this.

He has said:

"In practice, financial institutions use the “average daily balance” to calculate the interest on deposits."

Yes  this is correct  

Nauseated has then stated that:

"The result returns a value with limits between the two formulas. "

Yes I can see how this could possibly be justified..   but I would like Nauseated to justify/discuss this statement. :)