If sum of three consecutive term of A.P. is 48 and product of first and third is 252 then find d?
To find the common difference (d) in an arithmetic progression (A.P.), we can use the given information about the sum and product of three consecutive terms.
Let's assume that the first term of the A.P. is 'a', and the common difference is 'd'. The three consecutive terms would be 'a', 'a + d', and 'a + 2d'.
Given:
The sum of the three consecutive terms is 48:
a + (a + d) + (a + 2d) = 48
3a + 3d = 48
Dividing by 3:
a + d = 16 ----(1)
The product of the first and third term is 252:
(a)(a + 2d) = 252
Expanding and simplifying:
a^2 + 2ad = 252 ----(2)
Now, we can solve equations (1) and (2) simultaneously to find the values of 'a' and 'd'.
Substituting equation (1) into equation (2):
(a^2 + 2ad) = 252
(a^2 + 2a(16 - a) = 252
a^2 + 32a - 2a^2 = 252
Rearranging terms and simplifying:
a^2 - 32a + 252 = 0
Now, we can solve this quadratic equation for 'a' using factoring, completing the square, or using the quadratic formula. In this case, we will use factoring:
(a - 12)(a - 21) = 0
This gives two possible values for 'a':
a - 12 = 0 ---> a = 12
a - 21 = 0 ---> a = 21
For each value of 'a', we can now substitute it back into equation (1) to find the corresponding value of 'd'.
When a = 12:
a + d = 16
12 + d = 16
d = 16 - 12
d = 4
When a = 21:
a + d = 16
21 + d = 16
d = 16 - 21
d = -5
Therefore, the common difference (d) could be either 4 or -5, depending on the value of 'a'.
So, there are two possible APs that satisfy the given conditions:
1) The first term (a) is 12, and the common difference (d) is 4.
2) The first term (a) is 21, and the common difference (d) is -5.
3/2 (2a + 2d) = 48
(a)(a+2d) = 252
a = 16-d, so
(16-d)(16-d + 2d) = 252
Now just solve for d