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What is the sum of the arithmetic sequence 3, 9, 15..., if there are 34 terms?
Solution:
Given, the arithmetic sequence is 3, 9, 15,.....
First term, a = 3
Common difference, d = 9 - 3
d = 6
We have to find the sum of 34 terms.
The sum of the n terms of arithmetic sequence is given by
\(s_{n}=\frac{n}{2}(a+l)\)
Where, n = number of terms
a = first term
l = last term
The n-th term of an arithmetic sequence is given by an = a + (n - 1)d
\(a_{34}=3+(34-1)(6)\)
\(a_{34}=3+(33)(6)\)
\(a_{34}=3+198\)
\(a_{34}=201\)
Now, a = 3, l = 201, n = 34
\(s_{34}=\frac{34}{2}(3+201)\)
\(s_{34}=17(204)\)
\(s_{34}=3468\)
Therefore, the sum upto 34 terms is 3468.
What is the sum of the arithmetic sequence 3, 9, 15..., if there are 34 terms?
Summary:
The sum of the arithmetic sequence 3, 9, 15..., if there are 34 terms is 3468.
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