In an arithmetic series, the sum of the 5th to the 8th terms inclusive is 114, whilst the sum of the 12th to 15th terms inclusive is 198. What is the 21st term?
I have got 99=2a +3d and 57= 2a+3d Im not sure what to do now
clearly both those equations cannot be true, since 99≠57.
The sum of T5 to T8 is
S8 - S4
= 8/2(2a+7d) - 4/2(2a+3d)
= 4a+22d = 114
Similarly,
S15-S11 = 15/2(2a+14d) - 11/2(2a+10d)
= 4a+50d = 198
Subtract those two to get 28d=84, or d=3
So, a=12
T21 = 12 + 20*3 = 72
To find the 21st term of the arithmetic series, we need to determine the common difference (d) and the first term (a).
Let's work step-by-step to find the values of d and a:
Step 1: Find the sum of the 5th to the 8th terms using the given information:
The sum formula for an arithmetic series is: Sn = (n/2)(2a + (n-1)d)
Where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
Given that the sum of the 5th to the 8th terms is 114, we have:
Sn = 114
n = 4 (since the 5th to 8th terms accounts for 4 terms)
Using the sum formula, we can substitute the values and form an equation:
114 = (4/2)(2a + (4-1)d)
Simplifying, we get: 114 = 2(2a + 3d)
Step 2: Find the sum of the 12th to the 15th terms using the given information:
Following the same logic, we have Sn = 198 and n = 4 (for the 12th to 15th terms).
Substituting the values into the sum formula, we have:
198 = (4/2)(2a + (4-1)d)
Simplifying, we get: 198 = 2(2a + 3d)
Step 3: Solve the system of equations:
Based on the equations we derived from steps 1 and 2, we can set them equal to each other and solve for a and d.
2(2a + 3d) = 2(2a + 3d)
Rearranging terms, we have: 4a + 6d = 4a + 6d
Since the equation is an identity (which holds true for any value of a and d), we cannot determine specific values for a and d. This means that there are multiple possible arithmetic series that satisfy the given conditions.
Without knowing the specific values of a and d, we cannot determine the 21st term without additional information.
To find the 21st term of the arithmetic series, we need to determine the common difference (d) and the first term (a) of the series.
You have correctly set up two equations using the formulas for the sum of an arithmetic series:
1) Sum of the 5th to 8th terms:
S(5-8) = 114 = 4a + 6d
2) Sum of the 12th to 15th terms:
S(12-15) = 198 = 4a + 6d
To find the values of a and d, we can either solve this system of equations or use a different approach.
Let's use a different approach called "method of differences":
1) We know that the sum of the 5th to 8th terms is 114, which means the average of these four terms is 114/4 = 28.5.
So, the average of the 5th and 8th terms will be 28.5, which is the arithmetic mean of a+4d and a+7d.
28.5 = (a+4d + a+7d)/2
Simplifying, we get: 57 = 2a + 11d
2) Similarly, the sum of the 12th to 15th terms is 198, so the average of these four terms is 198/4 = 49.5.
Thus, the average of the 12th and 15th terms will also be 49.5, which is the arithmetic mean of a+11d and a+14d.
49.5 = (a+11d + a+14d)/2
Simplifying, we get: 99 = 2a + 25d
Now, we have a system of linear equations:
Equation 1: 57 = 2a + 11d
Equation 2: 99 = 2a + 25d
Subtracting Equation 1 from Equation 2, we can eliminate the variable "a":
(99 - 57) = (2a + 25d - 2a - 11d)
42 = 14d
Dividing both sides by 14, we find:
d = 3
Substituting the value of d back into Equation 1 (or Equation 2), we can solve for "a":
57 = 2a + 11(3)
57 = 2a + 33
24 = 2a
a = 12
Now that we have found the values of a and d, we are ready to find the 21st term:
a(21) = a + 20d
a(21) = 12 + 20(3)
a(21) = 12 + 60
a(21) = 72
Therefore, the 21st term of the arithmetic series is 72.