If $n^{\text{th}}$  term of 24, 20, 16, 12, … is same as $n^{\text{th}}$  term of $-11,-8,-5,\cdots$  then $n=$

$a_n=24+\left(n-1\right)\left(-4\right)\cdots \cdots \cdots \left(1\right)$

Let $Q=-11,-8,-5,\cdots$       Here $a=-11,d=3$

$a_n=-11+\left(n-1\right)\left(3\right)\cdots \cdots \cdots \left(2\right)$

But $\left(1\right)=\left(2\right)$

$24-4n+4=-11+3n-3$

$28-4n=3n-14$

$7n=42\Rightarrow n=6$