Question

if the first term of an A.P is 2 and common difference is 4 then the sum of its first 40 terms is?​

Answer 1

Answer:

a=2

d=4

n=40

Sn=?

Sn=n/2×{2a+(n-1) d}

Sn= 40/2 ×{2×2+39×4}

Sn=20×160

Sn=320

therefore sum of 40 terms is 320

Answer 2

Answer:

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

2

n= 40 ; a= 2 ; d = 4

S40 = 40 [2×2 + ( 40 - 1) ×4 ]

2

= 20 [ 4 + ( 39× 4) ]

= 20 [ 4 + 156]

= 20 × 160

= 3200 ANS.

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