how many terms of the arithmetic sequence -5,-1,3,… must be added to reach the sum of 400
a = -5
d = 4
So,
n/2 (-10+4(n-1)) = 400
n = 16
To find out how many terms of the arithmetic sequence -5, -1, 3, ... must be added to reach the sum of 400, we need to use the formula for the sum of an arithmetic series.
The formula is:
Sn = (n/2)(a + l)
Where:
Sn is the sum of the series
n is the number of terms
a is the first term
l is the last term
In this case, we know that the first term (a) is -5 and the common difference is 4.
The last term (l) can be found using the formula for the nth term of an arithmetic sequence:
l = a + (n - 1)d
Substituting the values, we have:
l = -5 + (n - 1)4
l = -5 + 4n - 4
l = 4n - 9
Now we can substitute the values of a and l into the formula for Sn:
400 = (n/2)(-5 + 4n - 9)
Simplifying the equation:
800 = -5n + 4n² - 9n
800 = 4n² - 14n
Rearranging the equation:
4n² - 14n - 800 = 0
This is a quadratic equation, which can be solved by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula to find the values of n:
n = (-b ± √(b² - 4ac)) / 2a
In this case, a = 4, b = -14, and c = -800:
n = (-(-14) ± √((-14)² - 4(4)(-800))) / (2 * 4)
n = (14 ± √(196 + 12800)) / 8
n = (14 ± √(12996)) / 8
n = (14 ± 114) / 8
We have two possible solutions:
n = (14 + 114) / 8 = 128 / 8 = 16
n = (14 - 114) / 8 = -100 / 8 = -12.5
Since the number of terms cannot be negative, the solution is n = 16.
Therefore, we need to add 16 terms of the arithmetic sequence to reach the sum of 400.
To find out how many terms of an arithmetic sequence need to be added to reach a certain sum, we can use the formula for the sum of an arithmetic sequence:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn is the sum of the first n terms,
a is the first term of the sequence,
d is the common difference between terms,
and n is the number of terms.
In this case, we are given the arithmetic sequence -5, -1, 3,... and we want to find the number of terms (n) needed to reach a sum of 400.
First, let's identify the variables in the given sequence:
a = -5 (the first term)
d = -1 - (-5) = 4 (common difference)
Now we can rearrange the formula to solve for n:
Sn = (n/2)(2a + (n-1)d)
Since we know the sum (Sn) we want to reach is 400, we can substitute that value into the formula:
400 = (n/2)(2*(-5) + (n-1)*4)
Simplifying further:
400 = (n/2)(-10 + 4n - 4)
400 = (n/2)(4n - 14)
Multiply both sides of the equation by 2 to eliminate the fraction:
800 = (4n - 14)n
Distribute the n:
800 = 4n^2 - 14n
Rearrange the equation:
4n^2 - 14n - 800 = 0
Now we have a quadratic equation, we can solve for n using factoring, completing the square, or using the quadratic formula. In this case, we can solve it by factoring:
(n - 25)(4n + 32) = 0
Setting each factor equal to zero:
n - 25 = 0 or 4n + 32 = 0
Solve for n in each equation:
n = 25 or 4n = -32
n = -32/4 = -8
Since the number of terms of a sequence cannot be negative, we discard the -8 solution.
Therefore, the correct answer is n = 25. So, 25 terms of the arithmetic sequence -5, -1, 3,... must be added to reach the sum of 400.