The AP is 10, 7, 4, …, –62 with $d = -3$, last term $l = -62$.
The 11th term from the end = $l + (11-1)(-d)^{-1}$... Using: $n$th term from end $= l + (n-1) \times (-d)$
$$= -62 + (11-1)(3) = -62 + 30 = \mathbf{-32}$$
Answer: (c) –32
To find the $n$th term from the end of an AP, treat the last term as the first term and reverse the common difference: $T_n(\text{from end}) = l + (n-1)(-d)$. Here $l = -62$, $n = 11$, $-d = 3$, giving $-62 + 30 = -32$. Alternatively, find total number of terms first (25 terms), then the 11th from end is the 15th from the start: $a_{15} = 10 + 14(-3) = -32$.