Question

If the first term and common difference of an A.P are 2 and 4 respectively,  then the sum of its first 40 terms is​

Answer 1

Answer:

We know the formula:

Sn=2n[2a+(n−1)d]

Here, n=number of terms=40

          a=first term=2

          d=common difference=4

So, putting the above values in formula we get,

S40=240[2(2)+(40−1)4]=20[4+39×4]=20[4+156]=20×160=3200

∴ Sum of 40 terms=3200

Answer 2

Answer:

S40 = 3200

Step-by-step explanation:

a (first term) = 2

d (common difference) = 4

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

S40 = 40/2 [2× 2 + (40 - 1) 4]

= 20 [ 4 + (39×4)]

= 20 [ 4 + 156]

= 20 (160)

= 3200