Find the sum of the first 8th term of a linear sequence whose 1st term is 6 & whose last term is 46
Assuming the "last" term is the 8th term, then
S8 = 8/2 (6+46) = _____
and that's "sum of the first 8 terms" not "first 8th term." There is only one 8th term.
Well, to find the sum of a linear sequence, we can use the formula:
Sum = (n/2)(2a + (n-1)d)
Where:
- n is the number of terms in the sequence
- a is the first term
- d is the common difference between terms
In this case, we know the first term (a = 6) and the last term (46). To find the common difference, we can subtract the first term from the last term:
d = 46 - 6
d = 40
Now we need to find the number of terms. Since the sequence goes from the first term (6) to the last term (46), we can consider that 1 term was added. So, the number of terms is:
n = (last term - first term) / common difference + 1
n = (46 - 6) / 40 + 1
n = 41 / 40 + 1
n = 2.025 + 1
n = 3.025
Since we can't have a fraction of a term, let's take the floor value of n:
n = floor(3.025)
n = 3
Now, we can substitute these values into the sum formula:
Sum = (n/2)(2a + (n-1)d)
Sum = (3/2)(2 * 6 + (3-1) * 40)
Sum = (3/2)(12 + 2 * 40)
Sum = (3/2)(12 + 80)
Sum = (3/2)(92)
Sum = 138
So, the sum of the first 8 terms of this linear sequence is 138.
To find the sum of the first 8 terms of a linear sequence, we need to find the common difference between terms and use the formula for the sum of an arithmetic series.
To find the common difference (d):
last term = first term + (n - 1) * d,
where n is the number of terms in the sequence.
Given:
First term (a1) = 6
Last term (an) = 46
Number of terms (n) = 8
Using the formula:
an = a1 + (n - 1) * d
46 = 6 + (8 - 1) * d
46 = 6 + 7d
40 = 7d
d = 40/7
d = 5.71 (rounded to two decimal places)
Now that we know the common difference (d), we can use the formula for the sum of an arithmetic series:
Sn = (n/2) * (a1 + an)
Sn = (8/2) * (6 + 46)
Sn = 4 * 52
Sn = 208
Therefore, the sum of the first 8 terms of this linear sequence is 208.
To find the sum of the first n terms of a linear sequence, you can use the formula:
Sn = n/2 * (2a + (n-1)d)
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
In this case, we are given that the first term (a) is 6 and the last term is 46. To find the common difference (d), we can subtract the first term from the last term:
d = last term - first term
d = 46 - 6
d = 40
So, the common difference is 40.
Now, we need to find the value of n, which is the 8th term. Since we know the first term (a) is 6 and the common difference (d) is 40, we can use the formula for the nth term of an arithmetic sequence:
an = a + (n-1)d
Substituting the values, we have:
8th term = 6 + (8-1) * 40
8th term = 6 + 7 * 40
8th term = 6 + 280
8th term = 286
So, the 8th term of the sequence is 286.
Now, we can find the sum of the first 8 terms (n = 8) using the sum formula:
Sn = 8/2 * (2*6 + (8-1)*40)
Sn = 4 * (12 + 7 * 40)
Sn = 4 * (12 + 280)
Sn = 4 * 292
Sn = 1168
Therefore, the sum of the first 8 terms of the linear sequence is 1168.