What is the sum of the arithmetic sequence 3, 9, 15..., if there are 36 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 36 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

\(a_{36}=3+(36-1)(6)\)

\(a_{36}=3+(35)(6)\)

\(a_{36}=3+210\)

\(a_{36}=213\)

Now, a = 3, l = 213, n = 36

\(s_{36}=\frac{36}{2}(3+213)\)

\(s_{36}=18(216)\)

\(s_{36}=3888\)

Therefore, the sum upto 36 terms is 3888.

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What is the sum of the arithmetic sequence 3, 9, 15..., if there are 36 terms?

Summary:

The sum of the arithmetic sequence 3, 9, 15..., if there are 36 terms is 3888.