If an A.p. Fifth term exceeds twice second term by 1. The tenth term xeceeds twice fourth term by 3
* find first term and common difference
* find sum of first twenty-five terms.
a+4d = 2(a+d)+1
a+9d = 2(a+3d)+3
Now crank it out.
To find the first term and common difference of an arithmetic progression (A.P.), we need to use the given information.
Let's denote the first term as 'a' and the common difference as 'd'.
1) The fifth term exceeds twice the second term by 1:
The formula for the nth term of an A.P. is: a + (n - 1)d
So, the fifth term can be written as: a + (5 - 1)d, which is given to be equal to 2 times the second term plus 1: 2a + 1.
This gives us the equation:
a + 4d = 2a + 1 ---(eq. 1)
2) The tenth term exceeds twice the fourth term by 3:
The tenth term can be written as: a + (10 - 1)d, which is given to be equal to 2 times the fourth term plus 3: 2(a + 3d) + 3.
This gives us the equation:
a + 9d = 2a + 6d + 3 ---(eq. 2)
Now, we have two equations with two unknowns. We can solve them simultaneously to find the values of 'a' and 'd'.
Subtracting equation (1) from equation (2), we get:
9d - 4d = 2a + 6d - a
Simplifying it further, we have:
5d = a ---(eq. 3)
Substituting this value of 'a' into equation (1), we get:
5d + 4d = 2(5d) + 1
Simplifying it further, we have:
9d = 10d + 1
-d = 1
d = -1
Substituting this value of 'd' into equation (3), we get:
5(-1) = a
a = -5
Therefore, the first term (a) of the A.P. is -5 and the common difference (d) is -1.
To find the sum of the first twenty-five terms of the A.P., we can use the formula for the sum of an A.P.:
Sn = (n/2)(2a + (n - 1)d)
Here, n = 25, a = -5, and d = -1.
Plugging in these values, we get:
S25 = (25/2)(2(-5) + (25 - 1)(-1))
Simplifying this expression will give you the sum of the first twenty-five terms of the A.P.