Find the sum of the first 41 terms of the following series, to the nearest integer.
2, 8,14,...

5002

Step-by-step explanation:

The given series is an arithmetic progression with

a = the first term

d = common difference

n = number of terms

S = the sum of the nth term

Recall that

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

As such,

a = 2

d = 8 -2 = 14 - 8 = 6

n = 41

Hence

S = 41/2(2*2 + (41 - 1)6)

= 41/2 ( 4+ 240)

= 41 * 244/2

= 5002

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: