Let the common difference of the AP be d.
The first term is 14, so the second term is 14+d, and the third term is 14+2d.
The sum of the first 5 terms is given by (5/2)(2*14 + (5-1)d) = 5(14+2d) = 70+10d.
The sum of the first 10 terms is given by (10/2)(2*14 + (10-1)d) = 10(14+4d) = 140+40d.
We are given that the sum of the first 5 terms and the sum of the first 10 terms are equal in magnitude but opposite in sign. This can be written as:
|70+10d| = |140+40d|
There are two cases to consider:
Case 1: 70+10d = 140+40d
Subtracting 10d from both sides and subtracting 140 from both sides gives 70-140 = 40d-10d, or -70 = 30d. Solving for d gives d = -7/3.
Case 2: 70+10d = -(140+40d)
Expanding the right side of the equation gives 70+10d = -140-40d. Adding 40d to both sides and adding 140 to both sides gives 70+50d = -140. Subtracting 70 from both sides gives 50d = -210, and solving for d gives d = -21/5.
So we have two possible values for d: -7/3 and -21/5. To find the third term of the AP, we need to find which value of d gives a third term that is an integer.
The third term of the AP is 14+2d, so for -7/3, the third term is 14 + 2*(-7/3) = 14 - 14/3 = 28/3 - 14/3 = 14/3 = 4 2/3.
For -21/5, the third term is 14 + 2*(-21/5) = 14 - 42/5 = 70/5 - 42/5 = 28/5 = 5 3/5.
Therefore, the third term of the AP can be either 4 2/3 or 5 3/5.
The first term of an AP is 14 and the sum of the first 5 terms and the first 10 terms are equal in magnitude but opposite in sign. The third term of the AP is...
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