In the cases where

(i) a_{n}=3+4n

a_{1}=3+4×1=3+4=7

a_{2}=3+4×2=3+8=11

a_{3}=3+4×3=3+12=15

d=a_{2}-a_{1}=11-7=4

=a_{3}-a_{2}=15-11=4

Hence a_{n}=3+4n forms an A.P

a_{15}=3+4×15=3+60=63

Sum of 15 terms is

S_{15}= n(a_{1}+a_{15})/2=15 (7+63)/2=15×70/2=15×35=525

(ii) a_{n}=9-5n

a_{1}=9-5×1=4

a_{2}=9-5×2=-1

a_{3}=9-5×3=-6

d=a_{2}-a_{1}=-1-4=-5

=a_{3}-a_{2}=-6+1=-5

Hence a_{n}=9-5n forms an A.P

a_{15}=9-5×15=-66

Sum of 15 terms is

S_{15}=n (a_{15}-a_{1})/2=15 (4-66)/2=15×(-62)/2=15×(-31)=-465