The sum of first 12 terms of an arithmetic series is 186, and the 20th term is 83. What is the sum of first 40 terms?
You want
40/2 (2a+39d)
where
12/2 (2a+11d) = 186
a+19d = 83
Just solve those last two for a and d, then plug them into the first formula.
To find the sum of the first 40 terms of the arithmetic series, we need to find the common difference (d) and the first term (a1).
Let's start by finding the common difference (d):
We are given that the 20th term is 83, which can be written as:
a20 = a1 + (20 - 1)d
83 = a1 + 19d
We can also find the 12th term using the same formula:
a12 = a1 + (12 - 1)d
To find a12, we can rearrange the formula:
a1 = a12 - (12 - 1)d
Now, let's plug in the given values into these equations:
83 = a1 + 19d
186 = a12 - 11d
Next, we need to solve these two equations simultaneously to find the values of a1 and d.
First, let's multiply the first equation by 11 to eliminate d:
11(83) = 11a1 + 209d
913 = 11a1 + 209d
Now, subtract the second equation from the first equation:
913 - 186 = 11a1 + 209d - (a12 - 11d)
727 = 12a1 + 220d - a12
Since we have expressed a1 in terms of a12 and d, we can replace a1 in the equation above:
727 = 12(a12 - (12 - 1)d) + 220d - a12
727 = 12a12 - 12(12 - 1)d + 220d - a12
727 = 12a12 - 12d(12 - 1) + 220d - a12
727 = 12a12 - 12d(11) + 220d - a12
727 = 12a12 - 132d + 220d - a12
727 = -132d + 220d
727 = 88d
Now, solve for d:
d = 727/88
d ≈ 8.26
Now that we have found the value of d, we can substitute it back into either of the original equations to find a1. Let's use the equation a20 = a1 + 19d:
83 = a1 + 19(8.26)
83 = a1 + 156.94
a1 ≈ -73.94
We have found the values of a1 and d.
Now, to find the sum of the first 40 terms, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a1 + (n - 1)d)
Substituting the values we found:
S40 = (40/2)(2(-73.94) + (40 - 1)(8.26))
S40 = 20(-147.88 + 310.58)
S40 = 20(162.7)
S40 = 3254
Therefore, the sum of the first 40 terms of the arithmetic series is 3254.