For the given series 10, 7, 4, ..... (-62). Find the 11th term from the end.

-32

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

we know,

last term =a+(n–1)d

 –62=10+(n–1)(–3)

 –72=(n–1)(–3)

 n–1=24

n=25

The 11th term from the last term will be the 15th term from the start.

So, a15=a+(n−1)d

=10+(15–1)(–3)=10–42=–32

i.e., the 11th term from the last term is –32

The correct answer is option (2) : -32