Inactive Tutor answered 07/24/19
13 + 16 + 19 + 22 + . . . + 67
is 19 terms
a1 = 13
an = 67
an = a1 + d(n-1)
67 = 13 + 3(n-1)
67 = 13 + 3n - 3
57 = 3n
19 = n [67 is the 19th term]
S = (n/2)(a1 + an)
S = (19/2)(13+67)
S = (19/2)(80)
S = 760
Shelby C.
asked 07/23/19Given the Arithmetic series
13 + 16 + 19 + 22 + . . . + 67
What is the value of sum?
Inactive Tutor answered 07/24/19
13 + 16 + 19 + 22 + . . . + 67
is 19 terms
a1 = 13
an = 67
an = a1 + d(n-1)
67 = 13 + 3(n-1)
67 = 13 + 3n - 3
57 = 3n
19 = n [67 is the 19th term]
S = (n/2)(a1 + an)
S = (19/2)(13+67)
S = (19/2)(80)
S = 760
The sum of the first n terms of an arithmetic sequence is:
S = (n/2)(a1 + an)
where n is the number of terms, a1 is the first term, and an is the last term. In this case, a1 = 13 and an = 67. So to find S we just need to know n, the number of terms. The explicit formula for the sequence is
an = a1 + d·n where d = 3 = the common difference between terms.
an = a1 + d·n
67 = 13 + 3n
54 = 3n
18 = n
Now just plug a1 = 13, an = 67 and n = 18 into the equation for the sum S above and compute the answer.
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