Answer:
Hope This Helps
Step-by-step explanation:
The common difference (d) = 2
The first element (a) = 3.
So, the n-th term is
I = a + (n-1).d
= 3 + (n-1)×2
= 1 + 2n.
So the required sum is
= (n/2)(a + I)
= (n/2)( 3 + 1 + 2n)
= (n/2)(4 + 2n)
= n (n+2)
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