# Math Help - Sequnces and Series help(and on limit question)

1. ## Sequnces and Series help(and on limit question)

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

2. 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...

3. 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

4. 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 $f(x)\rightarrow0$.

5. 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 $f(x)\rightarrow0$

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

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

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