Question : Find the sum of 6 + 8 + 10 + 12 + 14 .................. + 40.
Option 1: 414
Option 2: 424
Option 3: 1600
Option 4: 400
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Correct Answer: 414
Solution : 6 + 8 + 10 + 12 + 14 ......... + 40 Sum of first n even natural numbers = n (n + 1) Add and subtract 6 in the series 6 + 6 + 8 + 10 + 12 + 14 .................. + 40 - 6 $\because$ (6 = 2 + 4) 2 + 4 + 6 + 8 + 10 + 12 ...................+ 40 - 6 Total number of terms from 2 to 40 (n) = $\frac{(40 - 2)}{2} + (1)$ = 20 (n is the total number of terms, 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 Now, 2 + 4 + 6 + 8 + 10..................+ 40 - 6 = 420 - 6 = 414 $\therefore$ The sum of the series will be 414. Hence, the correct answer is 414.
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