Find the sum of 6 + 8 + 10 + 12 + 14 __________ + 40.1. 4242. 4003. 16

1.	Find the sum of 6 + 8 + 10 + 12 + 14 __________ + 40.1. 4242. 4003. 16004. 414
Answer» Correct Answer - Option 4 : 414 Given : 6 + 8 + 10 +12 + 14 ......... + 40 Formula used : Sum of first n even natural numbers = n (n + 1) Calculations : Add and subtract 6 in the series 6 + 6 + 8 + 10 + 12 + 14 .................. + 40 - 6 2 + 4 + 6 + 8 + 10 + 12 ...................+ 40 - 6 (6 = 2 + 4) Total number of terms from 2 to 40 n = (40 - 2)/2 + (1) = 20 (n is the total number of term, 40 is the last term and 2 is the common difference in the series) Total number of terms(n) = 20 So, 2 + 4 + 6 + 8 + 10 ...................... 40 = 20 (20 + 1) ⇒ 420 2 + 4 + 6 + 8 + 10..................+ 40 - 6 = 420 - 6 ⇒ 414 ∴ The sum of the series will be 414

Find the sum of 6 + 8 + 10 + 12 + 14 __________ + 40.1. 4242. 4003. 16004. 414

Answer» Correct Answer - Option 4 : 414

Given :

6 + 8 + 10 +12 + 14 ......... + 40

Formula used :

Sum of first n even natural numbers = n (n + 1)

Calculations :

Add and subtract 6 in the series

6 + 6 + 8 + 10 + 12 + 14 .................. + 40 - 6

2 + 4 + 6 + 8 + 10 + 12 ...................+ 40 - 6 (6 = 2 + 4)

Total number of terms from 2 to 40

n = (40 - 2)/2 + (1) = 20 (n is the total number of term, 40 is the last term and 2 is the common difference in the series)

Total number of terms(n) = 20

So,

2 + 4 + 6 + 8 + 10 ...................... 40 = 20 (20 + 1)

⇒ 420

2 + 4 + 6 + 8 + 10..................+ 40 - 6 = 420 - 6

⇒ 414

∴ The sum of the series will be 414