
we know all it's doing is adding 6 over again to each term to get the next one, so then

now for the explicit one
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D7%5C%5C%20d%3D6%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D7%2B%28n-1%296%5Cimplies%20a_n%3D7%2B6n-6%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BExplicit%20Formula%7D%7D%7B%5Cstackrel%7Bf%28n%29%7D%7Ba_n%7D%3D6n%2B1%7D%20%5C%5C%5C%5C%5C%5C%20therefore%5Cqquad%20%5Cqquad%20f%2810%29%3D6%2810%29%2B1%5Cimplies%20f%2810%29%3D61)
Answer:
x = 7
Step-by-step explanation:
12x - 16 = 68
Add 16 to each side
12x - 16+16 = 68+16
12x =84
Divide by 12
12x/12 = 84/12
x = 7
Answer:
42 inches square
Step-by-step explanation:
x=6
108/12=9
9-3=6
12+12+9+9=42
33.2 blah blah blah 20 character 33.2
T(n)=6-1=5
that means all terms are 5 because n does not appear on the right hand side.
However, if T(n)=6n-1, then
first term = T(1)=6(1)-1=5,
but second term T(2)=6(2)-1=11...