Sequnces and Series help(and on limit question)

**Porcelain** · May 17th 2009, 11:07 AM

I need help with these four off my review sheet. I would love some explanations

1. Find the specified term of the arithmetic sequence:
t2=9, t5=21; 41=__

2. Find the specified term of the geometric sequence:
t2=8,t4=72; find t6=__

3. Find the sum of the arithmetic series. If there is no sum, state so.
50+48+46+ ... +10

4. Sorry im not sure how to type this question so try to bare with me lol
Lim x-->infinity (radical(1+x)-1)/(x)

**SENTINEL4** · May 17th 2009, 01:09 PM

Originally Posted by Porcelain

I need help with these four off my review sheet. I would love some explanations

1. Find the specified term of the arithmetic sequence:
t2=9, t5=21; 41=_

In one arithmetic sequence we have : next term = previous term + number
In your example we will have...
t1+x=t2
t2=9
t3=t2+x
t4=t3+x
t5=t4+x
t5=21

By doing the replacements we get :
21=t4+x $\Rightarrow$ 21=t3+x+x $\Rightarrow$ 21=t2+x+x+x $\Rightarrow$ 21=9+3x $\Rightarrow$ 3x=12 $\Rightarrow$ x=4
So we have t1=5 , t2=9 , t3=13 , t4=17 , t5=21 , t6=25 .... and we can find that t10=41

Originally Posted by Porcelain

2. Find the specified term of the geometric sequence:
t2=8,t4=72; find t6=__

In geometric sequence we have : next term=previous term*number (say y). Here we have :
t2=8
t3=8*y
t4=t3*y
Working like before we find $y^2=9$ $\Rightarrow$ y=3 or y=-3

Now you can find t6.

Hope this is what you want...

**SENTINEL4** · May 17th 2009, 01:40 PM

Originally Posted by Porcelain

3. Find the sum of the arithmetic series. If there is no sum, state so.
50+48+46+ ... +10

You can find the sum from
$\sum=\frac{(a1+an)*n}{2}$ , where a1=first term, an=last term, n=number of terms

**VonNemo19** · May 17th 2009, 03:25 PM

Check this out

$\lim_{x\toinfty}\frac{\sqrt{1+x}-1}{x}=\lim_{x\toinfty}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}=\lim_{x\toinfty }\frac{x}{x(\sqrt{1+x}-1)}=$

$=\lim_{x\toinfty}\frac{1}{\sqrt{1+x}+1}$

You can see that the function is only defined for $x\geq{-1}$ and that as $x$ increases without bound, it is implied that $f(x)\rightarrow0$ .

SORRY ABOUT THE ERROR GUYS!

**VonNemo19** · May 17th 2009, 03:29 PM

Check this out

$\lim_{x\to\infty}\frac{\sqrt{1+x}-1}{x}=\lim_{x\to\infty}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}=\lim_{x\to\inft y}\frac{x}{x(\sqrt{1+x}-1)}=$

$=\lim_{x\to\infty}\frac{1}{\sqrt{1+x}+1}$

You can see that the function is only defined for $x\geq{-1}$ and that as $x$ increases without bound, it is implied that $f(x)\rightarrow0$

**Plato** · May 17th 2009, 03:42 PM

Originally Posted by Porcelain

3. Find the sum of the arithmetic series. If there is no sum, state so.
50+48+46+ ... +10

Recall that $\sum\limits_{k = 1}^n k = \left[ {\frac{{n(n + 1)}}{2}} \right]$ .

$\sum\limits_{k = 5}^{25} {2k} = 2\sum\limits_{k = 5}^{25} k = 2\left[ {\sum\limits_{k = 1}^{25} k - \sum\limits_{k = 1}^4 k } \right] = 2\left[ {\frac{{25(26)}}{2} - \frac{{4(5)}}{2}} \right]$

**Porcelain** · May 19th 2009, 05:29 PM

thanks everyone i think this will be my first 100 in this class

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