Question

if frist term of AP is 2 and common difference is 4 then sum of its 40 term is
1. 3200
2. 1600
3. 200
4. 2800
​

Answer 1

Answer:

1.3200

Step-by-step explanation:

S40=40/2[2a+(n-1)d]

= 20[2(2)+(40-1)4]

= 20[4+156]

=20×160

=3200

Answer 2

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

a=2,d=4,n=40

S40=40/2[2*2+(40-1)4]

=20[4+39*4]

=20[4+156]

=20[160]

=3200