Arithmetic progression. $a_3 + a_6 = -20$, $S_6 = -72$. Find $a_{11}$.

Question

Arithmetic progression. $a_3 + a_6 = -20$, $S_6 = -72$. Find $a_{11}$.

Formula for finding nth term is $a_n = a_1 + (n-1) * d$.

Formula for finding sum is $S_n = \frac{a_1 + a_n}{2} * n$.

I am trying to solve this for couple hours already. Farthest I went is that $a_1 = a_3 - 4$ by changing $a_3 + a_6 = -20$ to $a_6 = -20 - a_3$ and then pluging this into sum formula to find $a_1$.

Tried to find d by using one arithmetic progression property: $b = \frac{a + c}{2} $ to find $a_2$ and then just do usual $d = a_2 - a_1$ , but in the end I still have so many variables left.

Answer 1

$$\begin{cases} a_3+a_6=-20 \\ 6\frac{a_1+a_6}{2}=-72 \end{cases}$$ $$\begin{cases} a_1+2d+a_1+5d=-20 \\ a_1+a_1+5d=-24 \end{cases}$$

$$\begin{cases} a_1=-17 \\ d=2 \end{cases}$$ Then $a_{11}=a_1+10d=-17+10\cdot2=3$

Answer 2

Formula for finding nth term is $a_{n}=a_{k}+(n−k)d$, can you solve with this hint ?