In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20,.........,50.

Answer 1

an= a1 + (n-1)d where d= the common difference = 3 and a1 = 5

for n=11

a11 = 5 + (11-1)(3) = 5 + 30 = 35

a6 = 20

a7= 23

a8 = 26

a9 = 29

a10 = 32

a11 = 35

a16 =50 = 5+(16-1)3 = 5+15(3) = 5+45