I'm trying to solve this sequence but I can't find anything related to my exact problem.
Given the following:
$3$a3 + $6$a12 = 81
Specifically, I'm trying to find the 9th element [a9] as well as d - common difference.
I know of the formula
an = a1+(n-1)d
But not sure on how to implement it.
What would be a good way of approaching this problem?
Since $a_3 = a+2d$ and $a_{12} = a+ 11d $ , we get:
$$\begin{align}3a_3 + 6a_{12} &= 81 \\ 3a +6d +6a + 66d &= 81 \\ 9a+72d &= 81 \\ a+8d &= 9 \\ \color{blue}{\boxed{a_9}} &= 9\end{align}$$
The question doesn't provide enough information to find a unique value of $a$ or $d$.
With $$a_3=a_1+2d$$ and $$a_{12}=a_1+11d$$ you will get $$3(a_1+2d)+6(a_1+11d)=81$$
