First term of an ap is twice it's common difference. If the sum of the first two term of the AP is130. Find the first term of the ap.
a = 2d
a + a+d = 130
so,
2a+d = 130
4a+2d = 260
4a+a = 260
a = 52
Let's assume that the first term of the AP is 'a' and the common difference is 'd'.
According to the given information, the first term is twice the common difference, so we can write:
a = 2d .....(1)
We are also given that the sum of the first two terms of the AP is 130, so we can write:
a + (a + d) = 130 .....(2)
Substituting the value of 'a' from equation (1) into equation (2), we get:
2d + (2d + d) = 130
5d = 130
d = 130/5
d = 26
Now, substituting the value of 'd' back into equation (1), we get:
a = 2(26)
a = 52
Therefore, the first term of the AP is 52.
To find the first term of an arithmetic progression (AP), we'll use the given information. Let's denote the first term of the AP as 'a', and the common difference between the terms as 'd'.
Given that the first term is twice the common difference, we can write this as:
a = 2d ----(1)
The sum of the first two terms of the AP is given as 130. In an arithmetic progression, the sum of the first two terms can be calculated using the formula:
Sum = (n/2) * (2a + (n-1)d)
In this case, we have the sum as 130, and the first term 'a'. We need to determine 'd'.
Plugging in the values, the equation becomes:
130 = (2/2) * (2a + (2-1)d)
130 = a + d
Now we have a system of equations:
a = 2d ----(1)
130 = a + d
We can solve for 'a' and 'd' by substituting equation (1) into equation (2):
130 = 2d + d
130 = 3d
Dividing both sides of the equation by 3, we get:
d = 130/3
d = 43.33 (approx.)
Now, we have the value of 'd'. We can substitute this value back into equation (1) to find 'a':
a = 2d
a = 2 * 43.33
a = 86.67 (approx.)
Therefore, the first term of the arithmetic progression is approximately 86.67.