In the cases where
(i) an=3+4n
a1=3+4×1=3+4=7
a2=3+4×2=3+8=11
a3=3+4×3=3+12=15
d=a2-a1=11-7=4
=a3-a2=15-11=4
Hence an=3+4n forms an A.P
a15=3+4×15=3+60=63
Sum of 15 terms is
S15= n(a1+a15)/2=15 (7+63)/2=15×70/2=15×35=525
(ii) an=9-5n
a1=9-5×1=4
a2=9-5×2=-1
a3=9-5×3=-6
d=a2-a1=-1-4=-5
=a3-a2=-6+1=-5
Hence an=9-5n forms an A.P
a15=9-5×15=-66
Sum of 15 terms is
S15=n (a15-a1)/2=15 (4-66)/2=15×(-62)/2=15×(-31)=-465