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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