Answer:
N=11
Step-by-step explanation:
To find the value of "n" such that the sum of the first "n" natural numbers is 66, we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by:
Sum = (n/2) * (first term + last term)
In this case, the first term is 1 and the last term is "n". So we have:
66 = (n/2) * (1 + n)
To solve for "n", we can rearrange the equation and solve for "n" as follows:
66 = (n/2) * (1 + n)
66 = (n/2) + (n^2/2)
132 = n + n^2
n^2 + n - 132 = 0
Now we can factor the quadratic equation or use the quadratic formula to solve for "n". Factoring the equation, we get:
(n + 12)(n - 11) = 0
Setting each factor equal to zero, we find two possible values for "n":
n + 12 = 0 or n - 11 = 0
Solving for "n" in each case, we get:
n = -12 or n = 11
However, since we are looking for the value of "n" for the sum of the first "n" natural numbers, we can discard the negative value. Therefore, the value of "n" such that the sum of the first "n" natural numbers is 66 is:
n = 11