If sum of three consecutive term of A.P. is 48 and product of first and third is 252 then find d?
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To solve this problem, we can use the formula for the sum of the first n terms of an arithmetic progression (A.P.):
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.
Let's break down the given problem using this formula:
1. We are given that the sum of three consecutive terms is 48, so we can say Sn = 48 and n = 3.
Using the formula, we get:
48 = (3/2)(2a + 2d)
2. We are also given that the product of the first and third terms is 252, so we can say a(a + 2d) = 252.
3. We have two equations now:
48 = (3/2)(2a + 2d) (Equation 1)
a(a + 2d) = 252 (Equation 2)
To find the common difference d, we need to solve these two equations simultaneously.
Let's solve Equation 2 for a:
a(a + 2d) = 252
a^2 + 2ad = 252
a^2 = 252 - 2ad
a^2 = 252 - 2ad
Substitute this value of a in Equation 1:
48 = (3/2)(2a + 2d)
Divide both sides by (3/2):
32 = 2a + 2d
Now, substitute the value of a^2 from Equation 2:
32 = 2(252 - 2ad) + 2d
32 = 504 - 4ad + 2d
Rearrange the equation:
2d - 4ad = 504 - 32
-4ad + 2d = 472
Factor out d:
d(-4a + 2) = 472
Simplify:
-4a + 2 = 472/d
To solve for d, we need to find the value of a.
Let's solve Equation 2 for a:
a(a + 2d) = 252
Using the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / 2
In this case, a = 1, b = 0, and c = -252 (since a^2 + 2ad - 252 = 0).
a = (-0 ± √(0^2 - 4(1)(-252))) / 2
a = ± √(0 + 1008)) / 2
a = ± √(1008) / 2
a = ± 3√(28) / 2
Now that we have the values of a, we can substitute them into Equation 1 to find d.