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 a_n = 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.

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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.