Question

Find the sum of 15th term of each of the following sequence having nth term. (i) an=3+4n

Answer 1

an=3+4n
put n=1
a1=7
put a2=3+8=11
hence common difference (d)=11-7=4
and first term (a1)=7
now
Sn=n/2{2a+(n-1)d}. use this formula
hence,
S15=15/2{7 x 2 +(15-1)x4}=15(7+28)
=15 ×35 =525

Answer 2

1st term will be 3+4(1) =7
2nd term =3+4(2)=11
3rd=3+4(3)=15
so you can see that it is arithmetic progression as common difference is same
so it's sum formula is n/2(a+l)
where a is 1st term and l is last
l=3+4(15)=63
sum=15/2(63+7)
=525