Question

If first term of an A.P. is 2 & common difference is 4, then sum of its first 40 terms
is_____.

Answer 1

Answer:

3200

Step-by-step explanation:

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

    = 40[2×2 + (40-1)×4]/2

    = 20 [ 4 + 39×4 ]

    = 20 × [ 4 + 156 ]

    = 20 × 160

    = 3200

Answer 2

Answer:

The answer is 3160

Step-by-step explanation:

Given that a=2 and d=4

then An=A40=a+(n-1)d=a+39d

So A40=2+39 x 4 = 158

So Sn(sum of terms)=S40=n/2(a+An)

so n/2(a+An)=40/2(2+156)=3160

Hope it helps