The sum of the first and fourth terms of an A.P is 2 and their square is 20.how to find the sum of the first 8 terms?
use your usual formulas. You know that
a + a+3d = 2
a^2 + (a+3d)^2 = 20
You will get two solutions, and then you find
8/2 (2a+7d)
To find the sum of the first 8 terms of an Arithmetic Progression (A.P), we need to follow these steps:
Step 1: Determine the value of the first term (a) and the common difference (d) of the A.P.
Step 2: Use the given information to form two equations using the sum and square of the first and fourth terms.
Step 3: Solve the equations to find the values of a and d.
Step 4: Use the formula for the sum of the first n terms of an A.P to find the sum of the first 8 terms.
Let's solve the problem step-by-step:
Step 1: Determine the value of the first term (a) and the common difference (d) of the A.P.
Let the first term of the A.P be a, and the common difference be d.
Step 2: Use the given information to form two equations using the sum and square of the first and fourth terms.
Given that the sum of the first and fourth terms is 2, we have:
a + (a + 3d) = 2 ...(Equation 1)
Also, given that the square of the first and fourth terms is 20, we have:
a^2 + (a + 3d)^2 = 20 ...(Equation 2)
Step 3: Solve the equations to find the values of a and d.
First, simplify Equation 2:
a^2 + (a^2 + 6ad + 9d^2) = 20
2a^2 + 6ad + 9d^2 = 20 ...(Equation 3)
Now, substitute the value of a from Equation 1 into Equation 3:
2(2 - 3d)^2 + 6(2 - 3d)d + 9d^2 = 20
Expanding the equation and simplifying, we get:
4 - 12d + 9d^2 + 12d - 18d^2 + 9d^2 = 20
-9d^2 - 3d^2 + 12d - 12d + 4 = 20
-12d^2 + 4 = 20
-12d^2 = 16
d^2 = -16/12
d^2 = -4/3
d = ± √(-4/3) (Since d must be real)
As d cannot be an imaginary number, we discard the negative root.
So, d = √(-4/3)
Now, substituting the value of d into Equation 1, we can find the value of a:
a + (a + 3√(-4/3)) = 2
2a + 3√(-4/3) = 2
2a = 2 - 3√(-4/3)
2a = 2 - 3√(-1) √(4/3)
2a = 2 - 3i√(4/3)
2a = 2 - 3i(2/√3)
2a = 2 - (6i/√3)
2a = 2 - (2i√3)
So, the value of a is 2 - (2i√3).
Step 4: Use the formula for the sum of the first n terms of an A.P to find the sum of the first 8 terms.
The formula for the sum of the first n terms of an A.P is given by:
Sn = (n/2)(2a + (n-1)d)
Substituting the values of a and d into the formula and finding the sum of the first 8 terms:
S8 = (8/2)(2(2 - (2i√3)) + (8-1)√(-4/3))
Simplifying, we have:
S8 = 4(4 - 4i√3 + 7√(-4/3))
S8 = 16 - 16i√3 + 28√(-4/3)
So, the sum of the first 8 terms of the A.P is 16 - 16i√3 + 28√(-4/3).
To find the sum of the first 8 terms of an arithmetic progression (A.P.), we need to first find the common difference (d) and the first term (a).
Given that the sum of the first and fourth terms is 2, we can write it as:
a + (a + 3d) = 2 --(1)
Also, the square of the sum of these terms is 20:
(a + a + 3d)^2 = 20 --(2)
Simplifying equation (2):
(2a + 3d)^2 = 20
4a^2 + 12ad + 9d^2 = 20
Next, we need to solve equations (1) and (3) simultaneously to find the values of a and d.
Let's substitute the value of (a + 3d) from equation (1) into equation (3):
a + (a + 3d) = 2 --(1)
(2a + 3d)^2 = 20 --(3)
Substituting (1) into (3):
(2a + 3d)^2 = 20
(2a + 3d)^2 = 20
4a^2 + 12ad + 9d^2 = 20
Expand the square:
4a^2 + 12ad + 9d^2 = 20
Simplifying further:
4a^2 + 12ad + 9d^2 - 20 = 0
Divide all the terms by 4 to simplify:
a^2 + 3ad + (9d^2/4) - 5 = 0 --(4)
Now, we have a quadratic equation in terms of 'a'. To solve it, we can either factorize or use the quadratic formula.
We can see that equation (4) does not factorize easily, so let's solve it using the quadratic formula:
a = (-3d ± √(9d^2 - 4(9d^2/4)(-5))) / 2
Simplifying:
a = (-3d ± √(9d^2 + 45d^2)) / 2
a = (-3d ± √(54d^2)) / 2
a = (-3d ± √(6^2 * d^2)) / 2
a = (-3d ± 6d) / 2
a = (3d - 6d) / 2 or a = (-3d - 6d) / 2
a = -3d or a = -9d
We have two possible values of 'a' based on the equation:
Case 1: a = -3d --(5)
Case 2: a = -9d --(6)
Now, let's consider each case and calculate the sum of the first 8 terms of the A.P.
Case 1: a = -3d
The sum of the first 8 terms of an A.P. can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d)
where:
n = number of terms = 8
a = first term = -3d
d = common difference
Substituting the values in the formula:
S8 = (8/2)(2(-3d) + (8-1)d)
S8 = 4(-6d + 7d)
S8 = 4d
Therefore, for Case 1, the sum of the first 8 terms is 4d.
Case 2: a = -9d
Using the same formula:
Sn = (n/2)(2a + (n-1)d)
Substituting the values:
n = 8, a = -9d
S8 = (8/2)(2(-9d) + (8-1)d)
S8 = 4(-18d + 7d)
S8 = -44d
Therefore, for Case 2, the sum of the first 8 terms is -44d.
To find the actual sum of the first 8 terms of the A.P., we need to know the exact value of 'd'.