Question

How to find the sum of the sequence 3,9,15,...,1353?

Answer 1

This squence is in A.P and sum of A. p =n/2[2a+(n-1) d]

where a is first term, d is common difference

Answer 2

Step-by-step explanation:

sum = (n/2)*(2a+ (n-1)*d)

= (226/2)*(6+(225*6))

= (113)*(6+1350)

= 153228