Question

find the sum of 60 terms of the progressions 3,5,7,.....​

Answer 1

a=3

d=2

n=60

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

Sn=60/2(2×3+(60-1)×2)

=30×(6+118)

=30×124

=3720

Answer 2

a = 3

d = 5 - 3 = 2

n = 60

We know that,

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

S = 60/2 [2×3 +(60-1) × 2]

S = 30 [6 + 59×2]

S = 30 [6 + 118]

S = 30 × 124

S = 3720