Sum of terms of a PA

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THE Arithmetic Progression (PAN) it is a numerical sequence where the difference between two consecutive terms is always equal to the same value, a constant r.

For example, (1, 3, 5, 7, 9, 11, 13, 15) is an AP of ratio r = 2.

This type of sequence (PA) is very common and we may often want to determine the sum of all terms in the sequence. In the example above, the sum is given by 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64.

However, when the BP has many terms or when not all terms are known, it becomes more difficult to obtain this sum without using a formula. So, check out the formula for sum of terms of a PA.

Formula of sum of terms of a PA

THE sum of the terms of aArithmetic Progression can be determined by knowing only the first and last term of the sequence, using the following formula:

$\dpi{120} \small \mathbf{S_n = \frac{n.(a_1+a_n)}{2}}$

On what:

$\dpi{120} \mathbf{n}$ : number of PA terms;
$\dpi{120} \mathbf{a_1}$ : is the first term of the BP;
$\dpi{120} \mathbf{a_n}$ : is the last term of the PA.

Demonstration:

In demonstrating that the presented formula really allows to calculate the sum of the n terms of an AP, we must consider a very important property of the AP:

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Properties of a PA: the sum of two terms that are at the same distance from the center of a finite PA is always the same value, that is, constant.

To understand how this works in practice, consider the BP from the initial example (1, 3, 5, 7, 9, 11, 13, 15).

(1, 3, 5, 7, 9, 11, 13, 15) -> 1 + 15 = 16

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(1, 3, 5, 7, 9, 11, 13, 15) -> 3 + 13 = 16

(1, 3, 5, 7, 9, 11, 13, 15) -> 5 + 11 = 16

(1, 3, 5, 7, 9, 11, 13, 15) -> 7 + 9 = 16

Now, see that 16 + 16 + 16 + 16 = 4 x 16 = 64, which is the sum of the terms of this PA. Furthermore:

The number 16 can be obtained only through the first and last term 1+ 15 = 16.
The number 16 was added 4 times, which corresponds to half the number of terms in the sequence (8/2 = 4).

What happened is not a coincidence and goes for any PA.

In any PA, the sum of the equidistant terms will always be the same value, which can be obtained through ( $\dpi{120} \small \mathrm{a_1+ a_n}$ ) and as always are added every two values, in a sequence of $\dpi{120} \small \mathrm{n}$ terms, there will be ( $\dpi{120} \small \mathrm{a_1+ a_n}$ ) a total of $\dpi{120} \small \mathrm{\frac{n}{2}}$ times.