The 3rd and 7th terms of an A.P are -1 and 11 respectively. Fine the nth term and the numbers of which must be added to get a sum of 430
To find the nth term of an arithmetic progression (AP), we need to know the first term (a) and the common difference (d).
In this case, we are given the 3rd term (a3) as -1 and the 7th term (a7) as 11.
Step 1: Find the common difference (d)
Using the formula for the nth term of an AP: an = a + (n-1)d
We can set up two equations using a3 and a7 to solve for d.
a3 = a + 2d = -1
a7 = a + 6d = 11
Subtracting equation 1 from equation 2, we get:
4d = 12
d = 3
Step 2: Find the first term (a)
Using equation 1, we can substitute the value of d:
a + 2(3) = -1
a + 6 = -1
a = -7
Step 3: Find the nth term
Using the formula an = a + (n-1)d, we substitute the values of a and d:
an = -7 + (n-1)(3)
an = -7 + 3n - 3
an = 3n - 10
Therefore, the nth term of the arithmetic progression is 3n - 10.
Step 4: Find the numbers to be added to get a sum of 430
To find the sum of an arithmetic series (Sn), we can use the formula:
Sn = (n/2)(2a + (n-1)d)
We need to find the values of n and Sn.
Setting up the equation using the formula and the given sum:
430 = (n/2)(2(-7) + (n-1)(3))
Simplifying the equation:
430 = (n/2)(-14 + 3n - 3)
Expanding and rearranging the equation:
430 = (-14n + 3n^2 + 7n - 3)/2
Multiplying both sides by 2 to eliminate the fraction:
860 = -14n + 3n^2 + 7n - 3
Rearranging the equation:
3n^2 - 7n + 14n - 863 = 0
Solving the quadratic equation:
3n^2 + 7n - 863 = 0
Using the quadratic formula, we get:
n = (-b ± √(b^2 - 4ac)) / 2a
a = 3, b = 7, c = -863
n = (-7 ± √(7^2 - 4(3)(-863))) / 2(3)
n = (-7 ± √(49 + 10368)) / 6
n = (-7 ± √10417) / 6
n ≈ (-7 ± 102.07) / 6
Taking the positive value:
n ≈ (95.07) / 6
n ≈ 15.85
Therefore, the value of n is approximately 15.85.
To find the numbers that must be added to get a sum of 430, we can substitute the value of n into the sum formula:
Sn = (n/2)(2a + (n-1)d)
S15 ≈ (15/2)(2(-7) + (15-1)(3))
S15 ≈ (15/2)(-14 + 14(3))
S15 ≈ (15/2)(-14 + 42)
S15 ≈ (15/2)(28)
S15 ≈ 15(14)
S15 ≈ 210
To find the difference between the actual sum and the desired sum:
Difference = 430 - 210
Difference = 220
Therefore, the numbers that must be added to get a sum of 430 is approximately 220.
To find the nth term of an arithmetic progression (A.P.), we need two pieces of information: the common difference (d) and any term in the sequence.
Let's start by finding the common difference (d). We are given that the 3rd term is -1 and the 7th term is 11. This means that:
a + 2d = -1 (equation 1)
a + 6d = 11 (equation 2)
Here, a represents the first term, and d represents the common difference.
To solve this system of equations, we will perform a process called "elimination."
Multiplying equation 1 by 3, and equation 2 by 1, we get:
3(a + 2d) = 3(-1) (equation 3)
1(a + 6d) = 1(11) (equation 4)
Expanding and simplifying equations 3 and 4:
3a + 6d = -3 (equation 3)
a + 6d = 11 (equation 4)
Subtracting equation 4 from equation 3 (equation 3 - equation 4):
3a + 6d - (a + 6d) = -3 - 11
Simplifying:
3a - a = -14
2a = -14
Dividing both sides by 2:
a = -14/2
a = -7
Now that we have found the value of the first term (a = -7), we need to find the common difference (d). We can substitute the value of a into equation 2:
-7 + 6d = 11
Adding 7 to both sides:
6d = 11 + 7
6d = 18
Dividing both sides by 6:
d = 18/6
d = 3
So, the common difference is 3.
Now that we have the value of the first term (a = -7) and the common difference (d = 3), we can find the nth term.
The nth term formula for an arithmetic progression is:
nth term = a + (n - 1) * d
Substituting the values we found:
nth term = -7 + (n - 1) * 3
To find the number of terms that sum up to 430, we need to rearrange the nth term formula:
Sum = (n/2) * (2a + (n - 1) * d)
Rearranging and substituting the values:
430 = (n/2) * (-14 + (n - 1) * 3)
Expanding and simplifying:
430 = (n/2) * (-14 + 3n - 3)
430 = (n/2) * (3n - 17)
Multiplying both sides by 2:
860 = n(3n - 17)
Rearranging:
3n^2 - 17n - 860 = 0
Now we have a quadratic equation. We'll solve it using factoring or the quadratic formula.
Haule
a+2d = -1
a+6d = 11
subtract them:
4d = 12
d = 3
from a+2d = -1 , a = -7
term(n) = a+(n-1)d
= -7 + 3(n-1)
= -7 + 3n - 3
= 3n - 10
"the numbers of which must be added to get a sum of 430"
I will assume you mean, "how many terms are needed to have sum of 430" ?
sum(n) = (n/2)(first + last)
430 = (n/2)(-7 + 3n-10)
860 = n(3n - 17)
3n^2 - 17n - 860 = 0
(n - 20)(3n + 43) = 0
n = 20 or n = -43/3, the latter would be extraneous
you will need 20 terms