Find the sum of the first 40 terms of the arithmetic sequence. 19, 24, 29, 34
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Sn=sum of numbers
Sn=(n/2)*[2a1+(n-1)*d]
In this case:
n=40
a1=19
d=5
S40=(40/2)*[2*19+(40-1)*5]=20*(38+39*5)=20*(38+195)=20*233=4660
To find the sum of the first 40 terms of an arithmetic sequence, you can use the formula:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn is the sum of the first n terms,
n is the number of terms,
a is the first term, and
d is the common difference.
In this case, the first term (a) is 19 and the common difference (d) is 5 (24 - 19 = 5).
Let's plug in the values:
Sn = (40/2)(2(19) + (40-1)(5))
Simplifying,
Sn = (20)(38 + 39(5))
Sn = (20)(38 + 195)
Sn = (20)(233)
Sn = 4660
So, the sum of the first 40 terms of the arithmetic sequence is 4660.
To find the sum of the first 40 terms of an arithmetic sequence, we first need to identify the common difference (d) between the terms. In an arithmetic sequence, each term is obtained by adding or subtracting a constant value (d) to the previous term.
Looking at the given sequence: 19, 24, 29, 34, we can see that the common difference (d) is 24 - 19 = 5.
To find the sum of the first 40 terms (Sn), we can use the arithmetic series formula:
Sn = (n/2)(2a + (n - 1)d)
Where:
Sn is the sum of the first n terms of the sequence.
n is the number of terms.
a is the first term in the sequence.
d is the common difference.
In this case, n = 40, a = 19, and d = 5.
Substituting these values into the formula:
Sn = (40/2)(2(19) + (40 - 1)(5))
Sn = 20(38 + 39(5))
Sn = 20(38 + 195)
Sn = 20(233)
Sn = 4660
Therefore, the sum of the first 40 terms of the arithmetic sequence 19, 24, 29, 34 is 4660.