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April 30th, 2014, 07:05 PM   #1
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Arithmetic and geometric sequence problems Algebra 2

If anybody would be able to help me with either of the following problems that would be great (I don't know how to do a subscript so I'm just going to write "sub"):

Given the explicit formula for an arithmetic sequence find first 5 terms and 34th term: a sub n = -11 + 7n

If the first term of a geometric sequence is 2 and the common ratio r is 6, find the next 3 terms in the sequence and write the recursive formula. (For this one I already have the first part and just need the recursive formula)
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May 1st, 2014, 12:21 AM   #2
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Quote:
Originally Posted by FEB123 View Post
I don't know how to do a subscript so I'm just going to write "sub"
a_n or a_{n+1} (if there are more than one symbol in the subscript) in MATH tags
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Originally Posted by FEB123 View Post
Given the explicit formula for an arithmetic sequence find first 5 terms and 34th term: a sub n = -11 + 7n
From the given formula $\displaystyle a_n=-11+7n$ we could find $\displaystyle a_1$ and common difference d, because the formula in general is $\displaystyle a_n=a_1+(n-1)d$.
So, by rewriting the common formula we get $\displaystyle a_n=a_1-d+dn$.
Now we have $\displaystyle
\begin{eqnarray}
a_1-d&=&11 \\
dn & = & 7n
\end{eqnarray}$
Try to complete it yourself
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May 1st, 2014, 01:19 AM   #3
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Quote:
Originally Posted by FEB123 View Post

If the first term of a geometric sequence is 2 and the common ratio r is 6,

find the next 3 terms in the sequence and write the recursive formula.
$\displaystyle \dfrac{..}{..}\ \ a_1 = 2,\ r=6\\\;\\ \qquad a_n = a_1\cdot r^{n-1}\\\;\\ \qquad a_n= 2\cdot 6^{n-1}$

Last edited by aurel5; May 1st, 2014 at 01:25 AM.
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