Given:
a.) The sum of the first n consecutive even be found using S = n² + n where n > 2.
Let's determine the value of n at S = 30.
[tex]\text{ S = n}^2\text{ + n}[/tex][tex]\text{ 30 = n}^2\text{ + n}[/tex][tex]\text{ n}^2\text{ + n - 30 = 0}[/tex]Where, a = 1, b = 1 and c = -30
Using the quadratic formula:
[tex]\text{ n = x =}\frac{-\text{b }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{\text{ 2a}}\text{ = }\frac{\text{ -1 }\pm\text{ }\sqrt[]{(1)^2-4(1)(-30)}}{\text{ 2(1)}}[/tex][tex]\text{ = }\frac{\text{ -1 }\pm\text{ }\sqrt[]{1\text{ + 120}}}{2}\text{ = }\frac{\text{ -1 }\pm\text{ }\sqrt[]{121}}{2}[/tex][tex]\text{ n = }\frac{\text{ -1 }\pm\text{ 11}}{\text{ 2}}[/tex][tex]\text{ n}_1\text{ = }\frac{-1\text{ + 11}}{2}\text{ = }\frac{10}{2}\text{ = 5}[/tex][tex]\text{ n}_2\text{ = }\frac{-1\text{ - 11}}{\text{ 2}}\text{ = }\frac{\text{ -12}}{\text{ 2}}\text{ = -6}[/tex]Therefore, there are two possible values of n, it is 5 and -6.
The answer is letter A and B.
