The sum of first 6 terms of an Arithmetic progression is 6 the product of 2nd term and 5th term is -80 . Find the terms of A.P.
sum of first 6 terms
= (6/2)(2a + 5d) = 6
2a + 5d = 2 ---> a = (2-5d)/2 **
(a+d)(a+4d) = -89
a^2 + 5ad + 4d^2 = -80
sub ** into that
(2-5d)^2/4 + 5d(2-5d)/2 + 4d^2 = -80
times 4, the LCD
(2-5d)^2 + 10d(2-5d) + 16d^2 = -320
4 - 20d + 25d^2 + 20d - 50d^2 + 16d^2 + 320 = 0
-9d^2 + 324 = 0
d^2 = 36
d = ± 6
if d = 6, a = (2 - 30)/2 = -14
terms are: -14, -8, -2, 4, 10, 16, ....
you do the case when d = -6
Let's solve this step-by-step.
Step 1: Let's assume that the first term of the arithmetic progression is "a" and the common difference is "d".
Step 2: The sum of the first 6 terms of an arithmetic progression can be calculated using the formula: Sn = (n/2)(2a + (n-1)d), where "n" is the number of terms.
So, for the given problem, the sum of the first 6 terms is 6. We can set up the equation as:
6 = (6/2)(2a + (6-1)d)
Simplifying this equation, we get:
6 = 3(2a + 5d)
2a + 5d = 2 ----(Equation 1)
Step 3: The product of the 2nd term and the 5th term of an arithmetic progression can be calculated using the formula: T2 x T5 = (a + d)(a + 4d)
So, for the given problem, the product of the 2nd term and the 5th term is -80. We can set up the equation as:
-80 = (a + d)(a + 4d)
-80 = a^2 + 5ad + 4d^2 ----(Equation 2)
Step 4: Now we have a system of two equations containing two variables (a and d). We can solve this system of equations to find the values of a and d.
Solving Equations 1 and 2 simultaneously, we get:
From Equation 1:
2a + 5d = 2 ----(Equation 1)
From Equation 2:
a^2 + 5ad + 4d^2 = -80 ----(Equation 2)
We can solve this system of equations using substitution or elimination method.
Step 5: Let's solve using the substitution method. Solve Equation 1 for a in terms of d, then substitute it into Equation 2.
From Equation 1:
2a = 2 - 5d
a = (2 - 5d)/2
Substituting the value of a in Equation 2:
((2 - 5d)/2)^2 + 5((2 - 5d)/2)d + 4d^2 = -80
Simplifying the equation, we get a quadratic equation in terms of d.
Step 6: Solve the quadratic equation to find the values of d.
On solving the quadratic equation, we find that d = -4 or d = 2.
Step 7: Substitute the values of d back into Equation 1 to find the corresponding values of a.
For d = -4:
2a + 5(-4) = 2
2a - 20 = 2
2a = 2 + 20
2a = 22
a = 22/2
a = 11
For d = 2:
2a + 5(2) = 2
2a + 10 = 2
2a = 2 - 10
2a = -8
a = -8/2
a = -4
So, the terms of the arithmetic progression are 11, 7, 3, -1, -5, and -9 for d = -4, or
-4, -2, 0, 2, 4, and 6 for d = 2.
To find the terms of the arithmetic progression (A.P.), we can use the given information about the sum and the product of the terms. Here's how:
Let's assume that the first term of the A.P. is 'a', and the common difference is 'd'.
The sum of the first 6 terms of the A.P. can be calculated using the formula:
Sum = (n/2)(2a + (n-1)d), where 'n' is the number of terms.
Given that the sum is 6 and there are 6 terms, we have:
6 = (6/2)(2a + (6-1)d)
6 = 3(2a + 5d)
2a + 5d = 2 -----(Equation 1)
Now, let's consider the product of the 2nd term and the 5th term, which is -80. In an A.P., the 2nd term is 'a + d' and the 5th term is 'a + 4d'. Therefore, we have:
(a + d)(a + 4d) = -80
(a^2 + 5ad + 4d^2) = -80
a^2 + 5ad + 4d^2 + 80 = 0 -----(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) with two variables (a and d). We can solve these equations simultaneously to find the values of 'a' and 'd'.
From Equation 1, we can express 'a' in terms of 'd':
a = 2 - 5d -----(Equation 3)
Substituting the value of 'a' from Equation 3 into Equation 2, we get:
(2 - 5d)^2 + 5(2 - 5d)d + 4d^2 + 80 = 0
Expanding and simplifying the equation, we obtain:
29d^2 - 132d + 84 = 0
Now, we can solve this quadratic equation to find the value(s) of 'd'. Substituting the coefficients into the quadratic formula:
d = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 29, b = -132, and c = 84. Plugging in these values, we have:
d = (-(-132) ± √((-132)^2 - 4(29)(84))) / (2(29))
Simplifying further:
d = (132 ± √(17424 - 9756)) / 58
d = (132 ± √7668) / 58
Simplifying the square root, we get two possible values for 'd':
d₁ = (132 + √7668) / 58
d₂ = (132 - √7668) / 58
Now, substitute these values of 'd' back into Equation 1 to find the corresponding values of 'a'. Using Equation 3, we get:
a₁ = 2 - 5d₁
a₂ = 2 - 5d₂
Thus, if we calculated the values of 'd₁', 'd₂', 'a₁', and 'a₂', we would have the terms of the A.P.