Question:

1. Identify the first term and common difference:
$$a = 10, \quad d = 7-10 = -3.$$\
2. Let the total number of terms be $n$. The last term $a_n$ is given as $-62$.
$$a_n = a + (n-1)d = -62.$$\
3. Substitute $a$ and $d$:
$$10 + (n-1)(-3) = -62$$
$$-3(n-1) = -72$$
$$n-1 = 24 \Rightarrow n = 25.$$\
4. The 11th term from the last term means we move 10 positions towards the first term. Hence the required term is the $(n-10)^{\text{th}}$ term:
$$\text{Required term} = a_{n-10} = a_{25-10}=a_{15}.$$\
5. Compute the 15th term using the nth term formula:
$$a_{15} = a + (15-1)d = 10 + 14(-3) = 10 - 42 = -32.$$\
6. Therefore, the 11th term from the last term is $-32$.