Find the ${11}^{th}$ term from the last term (towards the first term) of the AP : $10,7,4,...,–62.$

The correct option is C: $-32$

Here, first term, $a=10,$ common difference, $d=7–10=–3,$ last term, $l=–62,$

Where, $l=a+(n–1)d$

To find the ${11}^{th}$ term from the last term, we will find the total number of terms in the AP.

So, $–62=10+(n–1)(–3)$

i.e., $–72=(n–1)(–3)$

i.e., $n–1=24$

or, $n=25$

So, there are $25$ terms in the given AP.

These are $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10},a_{11},a_{12},a_{13},a_{14},a_{15},a_{16},a_{17},a_{18},a_{19},a_{20},a_{21},a_{22},a_{23},a_{24},a_{25}$

The ${11}^{th}$ term from the last term will be the ${15}^{th}$ term from the start.

So, $a_{15}=a+(n-1)d$

$=10+(15–1)(–3)=10–42=–32$

i.e., the ${11}^{th}$ term from the last term is $–32.$