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DiKsa [7]
3 years ago
7

What is the sum of the first 51 consecutive odd positive integers?

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
4 0
We call:

a_{n} as the set of <span>the first 51 consecutive odd positive integers, so:

</span>a_{n} = \{1, 3, 5, 7, 9...\}

Where:
a_{1} = 1
a_{2} = 3
a_{3} = 5
a_{4} = 7
a_{5} = 9
<span>and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

</span>a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}
<span>
Then:

</span>a_{n} = a_{1} + (n-1)d
<span>
So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

</span>S_{k} = ( \frac{a_{1} +  a_{k}}{2}  ).k
<span>
Therefore, we need to find:
</span>a_{k} =  a_{51}  

Given that a_{1} = 1, then:

a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1

Thus:
a_{k} = a_{51} = 2(51)-1 = 101

Lastly:

S_{51} = ( \frac{1 + 101}{2} ).51 = 2601 

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