I'm learning context free grammars from languages.
Language ${L=\{{a}^{2i}\,{b}^j\,{c}^k\,|\,3i=j+k, i \gt 0\}}$
My guess is
$${S\rightarrow BA}$$ $${A\rightarrow Aaa|aa}$$ $${B\rightarrow bbbB\,|\,Bccc\,|\,bbAc\,|\,bAcc}$$
Since the word must start with an even number of a.
Am I right?
That doesn't seem to be correct since now you can produce an $a$ after $b$, which isn't allowed.
My attempt:
$S \to C_1c | B_1b $
$C_1 \to C_2c | B_1$
$C_2 \to aaC_3c | B_2$
$C_3 \to C_1c | B_3$
$B_3 \to B_1b | \epsilon$
$B_1 \to B_2b$
$B_2 \to aaB_3b$
The idea is that first you make $c$'s and then change the symbol to make $b$'s. You make two $a$'s at the left side every third time. For this you keep track with the lower index in B and C.
EDIT: I forgot the case that you can have zero $c$'s in which case you start to make immediately $b$'s. I also forgot that you need $i>0$ but now it's OK since you cannot make the string empty in the beginning.