Question

find the sum first 12 terms an A.P. whose nth term is given by an=3n+4​

Answer 1

Answer:

an= 3n+4

a1= 3(1)+4 = 3+4

a1=7

a12= 3(12)+4= 36+4

a12=40

Sn= n/2 (a+an)

S12=12/2 (a+a12)

=12/2(7+40)

=6(47)

S12=282

Answer 2

Answer:

n=12

a1=3+4=7

a2=3×2+4

a2=6+4=10

d=10-7=3

an=3n+4

a12=3×12+4

a12=36+4

a12=40

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

Sn=12/2[14+11(3)]

Sn=6[14+33]

Sn=6×47

Sn=282

ANSWER IS 282