Find the sum of the following arithmetic series.
13+11+9+...+(-25)
Someone help please. Thanks
well, in general for arithmetic sequence
sum from k = 0 to k = n-1
of a+kd
is
(n/2)[ 2a+(n-1)d ]
In this problem
a = 13
d = -2
now what is n?
13 + k(-2) = -25
-2 k = - 38
k = 19 for the last term
so n-1 = 19
n = 20
so we have
(20/2)[26 +19(-2)]
To find the sum of an arithmetic series, you can use the formula:
Sn = (n/2)(a₁ + aₙ)
Where Sn represents the sum of the series, n represents the number of terms, a₁ represents the first term, and aₙ represents the last term.
In this case, we need to find the sum of the arithmetic series:
13 + 11 + 9 + … + (-25)
First, we need to determine the values for n, a₁, and aₙ.
The series starts with a first term of 13 and ends with a last term of -25. To find the value of n, we can subtract -25 from 13 and divide by the common difference between each term, which is -2.
n = (-25 - 13) / -2
n = -38 / -2
n = 19
Now we know that there are 19 terms in the series.
Next, we can substitute the values into the formula:
Sn = (n/2)(a₁ + aₙ)
Sn = (19/2)(13 + (-25))
Sn = (19/2)(13 - 25)
Sn = (19/2)(-12)
Sn = -114
Therefore, the sum of the arithmetic series is -114.