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...
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.
Let's denote the common difference of the arithmetic progression (AP) as 'd'.
Given:
The first term of the AP is 14.
To find:
The third term of the AP.
We know that the sum of the first 'n' terms of an AP can be calculated using the formula: 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.
In this case, 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. Therefore, we can write the equation:
(-1)(5/2)(2*14 + (5-1)d) = (10/2)(2*14 + (10-1)d)
Simplifying the equation:
-5(28 + 4d) = 5(28 + 9d)
-140 - 20d = 140 + 45d
-65d = 280
Divide by -65 on both sides to solve for 'd':
d = -280 / -65
d ≈ 4.31
Now that we have the value of 'd', we can calculate the third term of the AP using the formula:
a3 = a + 2d
Substituting the given value of the first term 'a' as 14 and the value of 'd' as 4.31:
a3 = 14 + 2(4.31)
a3 ≈ 14 + 8.62
a3 ≈ 22.62
Therefore, the third term of the arithmetic progression (AP) is approximately 22.62.
To determine the third term of the arithmetic progression (AP), we need to find the common difference (d) between the terms.
Given information:
- First term (a₁) = 14
- Sum of the first 5 terms = - (Sum of the first 10 terms)
The sum of the first n terms of an AP is given by the formula:
Sn = (n/2) * (2a₁ + (n-1)d)
Since the sum of the first 5 terms is equal in magnitude but opposite in sign to the sum of the first 10 terms, we can write it as an equation:
(Sum of the first 5 terms) = -1 * (Sum of the first 10 terms)
=> (5/2) * (2a₁ + (5-1)d) = -1 * (10/2) * (2a₁ + (10-1)d)
Simplifying the equation:
(5/2) * (2a₁ + 4d) = -5 * (2a₁ + 9d)
5a₁ + 10d = -10a₁ - 45d
15a₁ = -55d
-3a₁ = 11d
We know that the first term (a₁) is 14. Substituting this value, we get:
-3 * 14 = 11d
-42 = 11d
d = -42/11
d ≈ -3.82
Now that we have the common difference (d), we can find the third term (a₃) using the formula:
a₃ = a₁ + 2d
Substituting the values:
a₃ = 14 + 2 * (-42/11)
a₃ = 14 - (84/11)
a₃ ≈ 6.45
Therefore, the third term of the AP is approximately 6.45.