What is the sum of the arithmetic sequence 3, 9, 15..., if there are 26 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 26 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

\(a_{26}=3+(26-1)(6)\)

\(a_{26}=3+(25)(6)\)

\(a_{26}=3+150\)

\(a_{26}=153\)

Now, a = 3, l = 153, n = 26

\(s_{26}=\frac{26}{2}(3+153)\)

\(s_{26}=13(156)\)

\(s_{26}=2028\)

Therefore, the sum upto 26 terms is 2028.

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What is the sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms?

Summary:

The sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms is 2028.