Find the 9th element of the arithmetic progression given the sum of 3 times 3rd term, and 6 times 12th term

Question

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

a_n = a₁+(n-1)d

But not sure on how to implement it.

What would be a good way of approaching this problem?

Answer 1

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 $d$.

Answer 2

With $$a_3=a_1+2d$$ and $$a_{12}=a_1+11d$$ you will get $$3(a_1+2d)+6(a_1+11d)=81$$