The fifth term of an A.P exceeds twice the second term by 1.The tenth term exceeds twice the fourth term by 3.
Find:1.the first term.
2.the common difference.
3.the sum of first twenty five terms.
Xn = a + (n-1)d
X5 = a + 4 d
X5 - 1 = 2 [ a + d ]
X10 - 3 = 2 [ a + 3 d ]
but X10 = a+9d
so
a + 4 d - 1 = 2 a + 2 d
a = 2 d - 1
a + 9 d - 3 = 2 a + 6 d
a = 3 d -3
combine
3 d - 3 = 2 d - 1
d = 4
You take it from there.
Let's start by finding the common difference using the given information.
1. The fifth term of an arithmetic progression (A.P) exceeds twice the second term by 1.
Let's assume the second term of the A.P is "a" and the common difference is "d". So, the fifth term can be expressed as: a + 4d.
According to the given information, we have the equation: a + 4d = 2a + 1.
Now, let's find the equation using the other given information.
2. The tenth term exceeds twice the fourth term by 3.
Using the same assumptions, the fourth term can be expressed as: a + 3d. The tenth term can be expressed as: a + 9d.
According to the given information, we have the equation: a + 9d = 2(a + 3d) + 3.
Now we have two equations:
a + 4d = 2a + 1 (equation 1)
a + 9d = 2(a + 3d) + 3 (equation 2)
To solve this system of equations, we can substitute equation 1 into equation 2.
Substituting a + 4d = 2a + 1 into a + 9d = 2(a + 3d) + 3, we get:
(2a + 1) + 9d = 2(a + 3d) + 3
2a + 1 + 9d = 2a + 6d + 3
9d - 6d = 3 - 1
3d = 2
d = 2/3
We have found the common difference, d, to be 2/3.
To find the first term, a, we can substitute d = 2/3 into equation 1 (a + 4d = 2a + 1).
a + 4(2/3) = 2a + 1
a + 8/3 = 2a +1
a - 2a = 1 - 8/3
-a = -5/3
a = 5/3
So, the first term, a, is 5/3.
Now, let's find the sum of the first twenty-five terms.
The sum of the first n terms of an arithmetic progression can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d)
Using the values we have:
n = 25 (the number of terms)
a = 5/3 (the first term)
d = 2/3 (the common difference)
Sn = (25/2)(2(5/3) + (25-1)(2/3))
Sn = (25/2)(10/3 + 24(2/3))
Sn = (25/2)(10/3 + 48/3)
Sn = (25/2)(58/3)
Sn = (25 * 58) / (2 * 3)
Sn = 725
The sum of the first twenty-five terms is 725.
To find the solution for this problem, we need to use the formulas of arithmetic progressions.
1. Finding the first term (a):
Let's assume that the common difference is 'd'.
According to the problem, the fifth term (a + 4d) exceeds twice the second term (a + 1d) by 1. So, we can write the equation as:
a + 4d = 2(a + 1d) + 1
Expanding the equation:
a + 4d = 2a + 2d + 1
Bringing like terms together:
a - 2a = -2d + 1d + 1
-a = -d + 1
Simplifying the equation:
a = d - 1
Therefore, the first term (a) is equal to the common difference (d) minus 1.
2. Finding the common difference (d):
Let's substitute the value of 'a' from the previous equation into the second equation from the problem.
The tenth term (a + 9d) exceeds twice the fourth term (a + 3d) by 3. So, we can write the equation as:
a + 9d = 2(a + 3d) + 3
Expanding the equation:
a + 9d = 2a + 6d + 3
Bringing like terms together:
a - 2a = 6d - 9d + 3
-a = -3d + 3
Simplifying the equation:
a = 3 - 3d
Now, we have two expressions for 'a', namely a = d - 1 and a = 3 - 3d.
Since both expressions are equal to 'a', we can set them equal to each other:
d - 1 = 3 - 3d
Bringing like terms together:
d + 3d = 3 + 1
4d = 4
Simplifying the equation:
d = 1
Therefore, the common difference (d) is equal to 1.
3. Finding the sum of the first twenty-five terms:
To find the sum of an arithmetic progression, we can use the formula:
Sum = (n/2)(2a + (n-1)d)
Given that the first term (a) is equal to -1 (from a = d - 1) and the common difference (d) is equal to 1, we can substitute these values into the formula:
Sum = (25/2)(2(-1) + (25-1)(1))
Simplifying the equation:
Sum = (25/2)(-2 + 24)
Sum = (25/2)(22)
Sum = 11(22)
Sum = 242
Therefore, the sum of the first twenty-five terms is 242.