Sum of first 12 terms of arithematic series is 186, 20th term is 186. Wat is sum of the first 40 terms
term(20)=186 ---> a+19d = 186
a = 186-19d
sum(12) = 6(2a + 11d) = 186
2a + 11d = 31
2(186-19d) + 11d = 31
-27d = -341
d = 341/27
a = 186-19(341/27) = -1457/27
sum(40) = 20(-1467/27+ 39(341/27) = -16041/27
better check my arithmetic, did not expect such weird numbers
To find the sum of the first 40 terms of an arithmetic series, we need to first find the common difference (d) and the first term (a1).
The formula to find the sum of an arithmetic series is:
Sn = (n/2) * (2a1 + (n-1)d)
Where:
Sn is the sum of the first n terms,
a1 is the first term of the series,
n is the number of terms in the series,
and d is the common difference.
We are given that the sum of the first 12 terms is 186. Therefore:
186 = (12/2) * (2a1 + (12-1)d)
186 = 6 * (2a1 + 11d)
31 = 2a1 + 11d ---(Equation 1)
We are also given that the 20th term is 186. Therefore:
a20 = a1 + (20-1)d
186 = a1 + 19d
a1 = 186 - 19d ---(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) to solve for a1 and d.
Substituting Equation 2 into Equation 1, we get:
31 = 2(186 - 19d) + 11d
31 = 372 - 38d + 11d
31 = 372 - 27d
27d = 372 - 31
27d = 341
d = 341/27
d ≈ 12.63
Substituting the value of d back into Equation 2, we can solve for a1:
a1 = 186 - 19d
a1 = 186 - 19(12.63)
a1 ≈ -20.87
Now we have obtained the values of a1 and d, we can substitute them into the formula for the sum of the first 40 terms of the arithmetic series:
S40 = (40/2) * (2a1 + (40-1)d)
S40 = 20 * (2(-20.87) + (39)(12.63))
Simplify the equation to find the sum:
S40 ≈ 20 * (-41.74 + 492.57)
S40 ≈ 20 * 450.83
S40 ≈ 9016.6
Therefore, the sum of the first 40 terms of the arithmetic series is approximately 9016.6.