find the sum of the first 24 terms of the arithmetic sequence, if the first term is 3 and the commom difference is 3.
Isnt there a formula for this? http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_ArithSeries.xml
sum(n) = n/2[2a + (n-1)d)
sum(24) = 12[6 + 23(3)]
= ....
To find the sum of the first 24 terms of an arithmetic sequence, use the formula for the sum of an arithmetic series:
Sum = (n/2)(2a + (n-1)d)
Where:
n = number of terms
a = first term
d = common difference
Given:
a = 3 (first term)
d = 3 (common difference)
n = 24 (number of terms)
Substituting the values into the formula:
Sum = (24/2)(2*3 + (24-1)*3)
= 12(6 + 23*3)
= 12(6 + 69)
= 12(75)
= 900
Therefore, the sum of the first 24 terms of the arithmetic sequence is 900.
To find the sum of the first 24 terms of an arithmetic sequence, you can use the formula for the sum of an arithmetic series. The formula is given by:
Sn = (n/2) * (2a + (n-1)d)
where:
Sn = the sum of the first n terms
n = the number of terms in the series
a = the first term
d = the common difference
In this case, the first term (a) is 3, the common difference (d) is 3, and we need to find the sum of the first 24 terms (n = 24). Plugging these values into the formula, we have:
S24 = (24/2) * (2*3 + (24-1)*3)
Let's simplify this expression step by step:
First, subtract 1 from 24:
S24 = (24/2) * (2*3 + 23*3)
Next, multiply 2 by 3:
S24 = (24/2) * (6 + 23*3)
Then, calculate 23 multiplied by 3:
S24 = (24/2) * (6 + 69)
Next, add 6 and 69 together:
S24 = (24/2) * 75
Now, divide 24 by 2:
S24 = 12 * 75
Finally, multiply 12 by 75:
S24 = 900
Therefore, the sum of the first 24 terms of the arithmetic sequence, where the first term is 3 and the common difference is 3, is 900.