Find the sum of the first 8 terms of a linear sequence whose first term is 6 and last term is 46
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To find the sum of the first 8 terms of a linear sequence, we first need to determine the common difference between each term.
Since the first term is 6 and the last term is 46, we can calculate the common difference using the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
Where:
an = nth term
a1 = first term
n = number of terms
d = common difference
Substituting the given values, we have:
46 = 6 + (8 - 1)d
Simplifying the equation:
46 = 6 + 7d
Subtracting 6 from both sides:
40 = 7d
Dividing by 7:
d = 40/7
Now that we know the common difference d = 40/7, we can use the formula for the sum of the first n terms of an arithmetic sequence:
Sn = (n/2)(a1 + an)
Substituting the known values:
S8 = (8/2)(6 + 6 + 40/7)
Simplifying:
S8 = 4(12 + 40/7)
To add these fractions, we need a common denominator:
S8 = 4(12 * 7/7 + (40/7))
S8 = 4(84/7 + 40/7)
S8 = 4(124/7)
Simplifying:
S8 = (4 * 124)/7
S8 = 496/7
Therefore, the sum of the first 8 terms of the given linear sequence is 496/7 or approximately 70.857.