Find the sum of the first eight terms of a linear sequence whose first term is 6 and last term is 46
tn=a+(n-1)d
T8=6+(8-1)d
46=6+(8-1)d
46=6+7d
7d=46-6
7d=40
d=40/7
d=5.7
sn=n/2(2a+<n-1>d)
=8/2(2times6(8-1)5.7)
=4(12+(39.9)
=4times51.9
=204
Is 208 the final answer?
sn=n/2(2a+<n-1>d)
If 46 is the 8th term, then
S8 = 8/2 (6+46) = 208
To find the sum of the first eight terms of a linear sequence, we can use the formula for the sum of an arithmetic series. The formula is given by:
S = (n/2)(2a + (n-1)d)
where:
S is the sum of the series,
n is the number of terms in the series,
a is the first term of the series, and
d is the common difference between consecutive terms.
In this case, the linear sequence starts with a = 6 and ends with the eighth term, which is 46. We need to find the value of d (the common difference) in order to calculate the sum.
The general formula for the nth term of an arithmetic sequence is given by:
An = a + (n-1)d
Using this formula, we can substitute the given values of the first term (6) and the last term (46) to solve for d.
46 = 6 + (8-1)d
46 = 6 + 7d
40 = 7d
d = 40/7
d ≈ 5.71
Now that we have found the common difference (d ≈ 5.71), we can substitute it into the sum formula to find the sum of the first eight terms:
S = (8/2)(2 * 6 + (8-1) * 5.71)
S = 4(12 + 7 * 5.71)
S ≈ 4(12 + 39.97)
S ≈ 4(51.97)
S ≈ 207.88
Therefore, the sum of the first eight terms of the given linear sequence is approximately 207.88.