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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Question : Find the mean proportion between $(6+\sqrt{8})$ and $(3-\sqrt{2})$.
Option 1: $2 \sqrt{12}$
Option 2: $\sqrt{14}$
Option 3: $(6-\sqrt{8})$
Option 4: $\sqrt{15}-7$
Question : If the sum of the measures of all the interior angles of a polygon is 1440$^\circ$, find the number of sides of the polygon.
Option 1: 8
Option 2: 12
Option 3: 10
Option 4: 14
Question : $6\frac{1}{4}$% of 1600 + $12\frac{1}{2}$% of 800 equals:
Option 1: 100
Option 2: 200
Option 3: 300
Question : If $4 x^2+y^2=40$ and $x y=6$, find the positive value of $2 x+y$.
Option 2: 6
Option 3: 5
Option 4: 4
Question : The value of $4 \div 12$ of $[3 \div 4$ of $\{(4-2) \times 6 \div 2\}]-2 \times 6 \div 8+3$ is:
Option 1: $4 \frac{1}{6}$
Option 2: $3 \frac{1}{3}$
Option 3: $2 \frac{1}{3}$
Option 4: $7 \frac{1}{6}$
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