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# What is the sum of the arithmetic sequence 3, 9, 15..., if there are 24 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 24 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

\(a_{24}=3+(24-1)(6)\)

\(a_{24}=3+(23)(6)\)

\(a_{24}=3+138\)

\(a_{24}=141\)

Now, a = 3, l = 141, n = 24

\(s_{24}=\frac{24}{2}(3+141)\)

\(s_{24}=12(144)\)

\(s_{24}=1728\)

Therefore, the sum upto 24 terms is 1728.

## What is the sum of the arithmetic sequence 3, 9, 15..., if there are 24 terms?

**Summary:**

The sum of the arithmetic sequence 3, 9, 15..., if there are 24 terms is 1728.

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