What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?
Schuyler, the sum of the 12 term aritmetic progression is 816
Schuyler, in order to solve your problem I had to turn it around (Gloria Estefan?). You said the last term was 13...I made it the first term. You said the common difference was a -10...I made it a +10.
Now I had to find the last term so I used this formula: L=a+(n-1)d so 13+(12-1)10
L=123
Then I used this formula:
Sn=n/2(a+L)
Sn=12/2(13+123)
Sn=6 Times 136
Sn=816
To find the sum of a 12-term arithmetic sequence, you need to use the formula for the sum of an arithmetic series:
Sn = (n/2) * (a1 + an)
Where:
- Sn represents the sum of the series
- n is the number of terms in the series
- a1 is the first term
- an is the last term
In this case, you are given that the last term (an) is 13 and the common difference is -10 in an arithmetic sequence of 12 terms. First, let's find the first term (a1) using the given information.
To find the first term, you can use the formula:
an = a1 + (n-1) * d
Where:
- n is the number of terms
- d is the common difference
Substituting the given values:
13 = a1 + (12-1) * (-10)
Simplifying the equation:
13 = a1 + (11) * (-10)
13 = a1 - 110
Adding 110 to both sides:
123 = a1
Now that we know the first term (a1), we can calculate the sum of the 12-term arithmetic sequence using the formula for the sum of an arithmetic series.
Sn = (n/2) * (a1 + an)
Substituting the given values:
Sn = (12/2) * (123 + 13)
Simplifying the expression:
Sn = 6 * (136)
Calculating the sum:
Sn = 816
Therefore, the sum of the 12-term arithmetic sequence is 816.