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storchak [24]
3 years ago
14

What is the sum of the first 37 terms of the arithmetic sequence?

Mathematics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

The sum of the first 37 terms of the arithmetic sequence is 2997.

Step-by-step explanation:

Arithmetic sequence concepts:

The general rule of an arithmetic sequence is the following:

a_{n+1} = a_{n} + d

In which d is the common diference between each term.

We can expand the general equation to find the nth term from the first, by the following equation:

a_{n} = a_{1} + (n-1)*d

The sum of the first n terms of an arithmetic sequence is given by:

S_{n} = \frac{n(a_{1} + a_{n})}{2}

In this question:

a_{1} = -27, d = -21 - (-27) = -15 - (-21) = ... = 6

We want the sum of the first 37 terms, so we have to find a_{37}

a_{n} = a_{1} + (n-1)*d

a_{37} = a_{1} + (36)*d

a_{37} = -27 + 36*6

a_{37} = 189

Then

S_{37} = \frac{37(-27 + 189)}{2} = 2997

The sum of the first 37 terms of the arithmetic sequence is 2997.

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Answer:

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Step-by-step explanation:

Given expression : \sum_{n=3}^{20}(n(n+1))

Solving further :

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\Rightarrow \sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n

So, \sum_{n=3}^{20}(n(n+1))=\sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n

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