the sum of the first 8 term of an arithmetic progresion is 220,if the third term is 17.Find the sum of the first 6 items
the third term is 17 ---> a+2d = 17
a = 17-2d
sum of first 8 terms = 220
---> (8/2)(2a + 7d) = 220
4(2a + 7d) = 220
2a + 7d = 55
2(17-2d) + 7d = 55
34 - 4d + 7d = 55
3d = 21
d = 7
then a = 3
sum(6) = 3(6 + 35) = 123
check:
terms are: 3 10 17 24 31 38 45 52 ..
sum(8) = 220
sum(6) = 123
all is good
Why did the arithmetic progression go to therapy?
Because it had a problem with its terms!
Now, let's solve this puzzle. We are given that the sum of the first 8 terms is 220. That means the average of those terms is 220 divided by 8, which is 27.5.
Since we know it's an arithmetic progression and the third term is 17, we can assume the common difference (d) between the terms is 10 (because the third term is the first term plus two times the common difference, so 17 = first term + 2 * d, which simplifies to d = 10).
Now, we're asked to find the sum of the first 6 terms, which we can denote as S6. We can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d)
Plugging in the values, we get:
S6 = (6/2)(2a + (6-1)d)
S6 = 3(2a + 5d)
S6 = 6a + 15d
But we don't know the first term (a), so we can't find the exact value of S6. However, we can make an approximation based on the average of the terms. Since the average is 27.5, we can assume that the sum of the first 6 terms is approximately 6 * 27.5, which is 165.
So, the sum of the first 6 terms is approximately 165. But don't take my word for it, do the math and enjoy your arithmetic journey!
To find the sum of the first 6 terms of an arithmetic progression, we can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d)
Where:
- Sn is the sum of the first n terms,
- a is the first term of the arithmetic progression,
- d is the common difference between the terms, and
- n is the number of terms.
Given that the sum of the first 8 terms (n = 8) is 220, and that the third term (n = 3) is 17, we can use this information to find the first term (a) and the common difference (d).
Step 1: Find the first term (a)
Using the formula for the sum of an arithmetic progression, we have:
220 = (8/2)(2a + (8-1)d)
220 = 4(2a + 7d)
55 = 2a + 7d
Step 2: Find the common difference (d)
Given that the third term is 17:
a + (3-1)d = 17
a + 2d = 17
Step 3: Solve the system of equations
We can now solve the system of equations consisting of Step 1 and Step 2.
We substitute 17 for a + 2d in the equation 55 = 2a + 7d:
55 = 2(17) + 7d
55 = 34 + 7d
7d = 55 - 34
7d = 21
d = 21/7
d = 3
Now, substitute the value of d into the equation a + 2d = 17:
a + 2(3) = 17
a + 6 = 17
a = 17 - 6
a = 11
We have found that the first term (a) is 11 and the common difference (d) is 3.
Step 4: Find the sum of the first 6 terms (n = 6)
Using the formula for the sum of an arithmetic progression, we have:
Sn = (n/2)(2a + (n-1)d)
S6 = (6/2)(2(11) + (6-1)(3))
S6 = 3(22 + 5(3))
S6 = 3(22 + 15)
S6 = 3(37)
S6 = 111
Therefore, the sum of the first 6 terms is 111.
To find the sum of the first 6 terms of an arithmetic progression, we need to know the common difference (d).
First, let's find the common difference (d):
Given that the third term is 17, we can use the arithmetic progression formula to find d:
a3 = a1 + (3-1)d
17 = a1 + 2d
We also know that the sum of the first 8 terms is 220:
S8 = (n/2)(2a1 + (n-1)d)
220 = (8/2)(2a1 + (8-1)d)
220 = 4(2a1 + 7d)
220 = 8a1 + 28d
Now we have two equations:
17 = a1 + 2d (Equation 1)
220 = 8a1 + 28d (Equation 2)
We can solve these equations simultaneously to find the values of a1 and d.
Subtracting Equation 1 from Equation 2:
220 - 17 = 8a1 + 28d - (a1 + 2d)
203 = 7a1 + 26d
Divide the resulting equation by 7:
203/7 = (7a1 + 26d)/7
29 = a1 + 3.71d (approximately)
Now, let's solve for a1:
29 = a1 + 3.71d
a1 = 29 - 3.71d
Substitute the value of a1 in Equation 1:
17 = 29 - 3.71d + 2d
17 = 29 - 1.71d
1.71d = 29 - 17
1.71d = 12
d = 12/1.71
d ≈ 7.02
Now that we have the value of d, we can find the sum of the first 6 terms:
S6 = (6/2)(2a1 + (6-1)d)
S6 = 3(2(29 - 3.71d) + 5d)
S6 = 3(58 - 7.42d + 5d)
S6 = 3(58 - 2.42d)
S6 = 3(58 - 2.42(7.02))
S6 = 3(58 - 16.9876)
S6 = 3(41.0124)
S6 = 123.0372 (approximately)
Therefore, the sum of the first 6 terms of the arithmetic progression is approximately 123.