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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 an = 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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