Answer 1

1/15, 1/12, 1/10, ...... , to 11 terms

For this A.P.,

a = 1/15

n =11

`d = a_2-a_1=1/12-1/15`

`=(5-4)/60 = 1/60`

We know that

`S_n = n/2[2a + (n -1)d]`

`S_11 = 11/2[2(1/15)+(11-1)1/60]`

`=11/2[2/5+10/60]`

`=11.2[2/15+1/6]=11/2[(4+5)/30]`

`=(11/2)(9/30)=33/20`