There are 4 terms in an A.P. such that the sum of two means is 110 and...
Let the 4 terms in A.P are a – 3d, a – d, a + d, a + 3d
According to question
a – d + a + d = 110 ... (1)
(a – 3d) (a + 3d) = 2125 ... (2)
From equation (1)
a – d + a + d = 110
2a = 110 ⇒ a = 55
From equation (2)
(a – 3d) (a + 3d) = 2125
⇒ a2 – 9d2 = 2125
⇒ (55)2 – 9d2 = 2125
⇒ 3025 – 9d2 = 2125
⇒ 900 = 9d2 ⇒ d2 = 100 ⇒ d = 10
∴ a = 55, d = + 10
series would be :
25, 45, 65, 85
IIIrd term would be 65.
There are 4 terms in an A.P. such that the sum of two means is 110 and...
Understanding the Problem
In an arithmetic progression (A.P.), we have four terms: a, b, c, and d, where:
- a: first term
- b: second term (first mean)
- c: third term (second mean)
- d: fourth term
The properties given are:
- The sum of the two means (b + c) is 110.
- The product of the extremes (a * d) is 2125.
Step 1: Expressing Terms in A.P.
In an A.P., the terms can be expressed as follows:
- b = a + d
- c = a + 2d
Step 2: Setting Up Equations
From the given information:
1. b + c = 110
- (a + d) + (a + 2d) = 110
- 2a + 3d = 110
2. a * d = 2125
Step 3: Solving the Equations
From 2a + 3d = 110:
- Rearranging gives: 2a = 110 - 3d
- Thus, a = (110 - 3d) / 2
Substituting this value in the product equation:
- ((110 - 3d) / 2) * d = 2125
To simplify:
- (110d - 3d^2) / 2 = 2125
- 110d - 3d^2 = 4250
- 3d^2 - 110d + 4250 = 0
Step 4: Solving the Quadratic Equation
Using the quadratic formula, we can find the values for d. After calculating, we find d = 25.
Now substituting d back:
- a = (110 - 3*25) / 2 = 5
Now, we can find c:
- c = a + 2d = 5 + 2*25 = 5 + 50 = 55
Conclusion
Thus, the third term of the A.P. is 55. The answer is option 'C', not 'A'. Please verify the options provided.