The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36,

(b) 6

$a=1, a_{n}=11, S_{n}=36$

$\because a_{n}=11$

$\Rightarrow a+(n-1) d=11$

$\Rightarrow 1+(n-1) d=11$

$\Rightarrow(n-1) d=10$   ....(1)

Also, $S_{n}=36$

$\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=36$

$\Rightarrow n\{2+10\}=72$   (Using (1))

$\Rightarrow n=6$