Originally Posted by
Soroban Hello, watcher!
You're expected to know this formula . . .
The $\displaystyle n^{th}$ term is: .$\displaystyle a_n \;=\;a + (n-1)d$
. . where $\displaystyle a$ is the first term, and $\displaystyle d$ is the common difference.
The $\displaystyle 4^{th}$ term is: .$\displaystyle a_4 \:=\:a + 3d$
The $\displaystyle 7^{th}$ term is: .$\displaystyle a_7 \;=\;a + 6d$
Their difference is 12: .$\displaystyle (a+6d)-(a+3d) \:=\:12 \quad\Rightarrow\quad 3d \:=\:12 \quad\Rightarrow\quad\boxed{d = 4}\;\;{\color{red}[1]}$
Their sum is -28: .$\displaystyle (a+6d)+(a+3d)\:=\:\text{-}28 \quad\Rightarrow\quad 2a+9d \:=\:\text{-}28\;\;{\color{red}[2]}$
. . Substitute [1] into [2]: .$\displaystyle 2a + 9(4) \:=\:\text{-}28 \quad\Rightarrow\quad\boxed{a \:=\:\text{-}32}$
Hence, the $\displaystyle n^{th}$ term is: .$\displaystyle a_n \;=\;-32 + (n-1)4 \quad\Rightarrow\quad {\color{blue}a_n \:=\:4n-36}$
The $\displaystyle 10^{th}$ term is: .$\displaystyle a_{10} \;=\;4(10)-36 \quad\Rightarrow\quad{\color{blue}a_{10}\;=\;4}$
The $\displaystyle 20^{th}$ term is: .$\displaystyle a_{20} \;=\;4(20)-36 \quad\Rightarrow\quad{\color{blue}a_{20}\;=\;44}$