Product of n terms of a geometric progression

The objective is to find a formula to calculate the product of the first terms of a geometric progression without needing to calculate them.

To do this, we will use the following property:

If we consider $n$ terms of a geometric progression, the product of two terms equidistant to the extremes is the same as the product of the extremes. In other words, the product of the first by the last term is the same as the product of the second by the penultimate term and so on, no matter how many terms we are considering in a geometric progression.

Example

Let's consider the geometric progression with the first term $a_{1} = 2^{0} = 1$ and ratio $r = 2.$

The first six terms are:

$a_{1} = 2^{0} = 1$ , $a_{2} = 2^{1} = 2$ , $a_{3} = 2^{2} = 4$ , $a_{4} = 2^{3} = 8$ , $a_{5} = 2^{4} = 16$ , $a_{6} = 2^{5} = 32.$

If we compute the product of the equidistant terms, we obtain:

$a_{1} \cdot a_{6} = 2^{0} \cdot 2^{5} = 2^{5}$ , $a_{2} \cdot a_{5} = 2^{1} \cdot 2^{4} = 2^{5}$ , $a_{3} \cdot a_{4} = 2^{2} \cdot 2^{3} = 2^{5} .$

Then the product of equidistant terms in the extremes is equal to the product of the extremes.

This is because the equidistant terms are obtained by increasing the first and reducing the last in the same proportion. Therefore, the product of these two factors must be the same as the product of the starting factors: the extremes.

Let $a_{1}$ and $a_{n}$ be the extremes, and $a_{1 + k}$ a term placed $k$ positions after the first one, and $a_{n - k}$ a term placed $k$ positions before the last one, we want to see that $a_{1} \cdot a_{n} = a_{1 + k} \cdot a_{n - k}$ .

As these terms are from a geometric progression, we know that:

$a_{1 + k} = a_{1} \cdot r^{1 + k - 1} = a_{1} \cdot r^{k}$ $a_{n - k} = a_{n} \cdot r^{n - k - n} = a_{n} \cdot r^{- k}$

And from here:

$a_{1 + k} \cdot a_{n - k} = (a_{1} \cdot r^{k}) \cdot (a_{n} \cdot r^{- k}) = (a_{1} \cdot a_{n}) \cdot (r^{k} \cdot r^{- k}) = a_{1} \cdot a_{n}$ that is what we wanted to prove.

Using this, we can see that the product $P_{n}$ of the $n$ first terms of a geometric progression is:

$P_{n} = \sqrt{(a_{1} \cdot a_{n})^{n}}$

Indeed, if $a_{1}, a_{2}, \dots, a_{n - 1}, a_{n}$ are the $n$ first terms, it will be $P_{n} = a_{1} \cdot a_{2} \cdot \dots \cdot a_{n - 1} \cdot a_{n}$ , or, $P_{n} = a_{n} \cdot a_{n - 1} \cdot \dots \cdot a_{2} \cdot a_{1}$ .

If we multiply both equalities member by member, we obtain:

$(P_{n})^{2} = (a_{1} \cdot a_{2} \cdot \dots \cdot a_{n - 1} \cdot a_{n}) (a_{n} \cdot a_{n - 1} \cdot \dots \cdot a_{2} \cdot a_{1}) =$

$= (a_{1} \cdot a_{n}) (a_{2} \cdot a_{n - 1}) \dots (a_{n - 1} \cdot a_{2}) (a_{n} \cdot a_{1})$

In the second member there appear $n$ brackets containing the product of two terms equidistant to the extremes that, as we have just seen, it is equal to the product of the extremes.

So:

$(P_{n})^{2} = (a_{1} \cdot a_{n}) (a_{1} \cdot a_{n}) \cdot \overset{n)}{\dots} \cdot (a_{1} \cdot a_{n}) (a_{1} \cdot a_{n}) = (a_{1} \cdot a_{n})^{n}$

And extracting the square root we obtain:

$P_{n} = \sqrt{(a_{1} \cdot a_{n})^{n}}$ that is what we wanted to see.

Example

To calculate the product of the first six multiples of $2$ , we notice that it is a geometric progression with the first term $a_{1} = 2^{0} = 1$ and ratio $r = 2$ .

So its general term is: $a_{n} = 2^{n - 1}$ , and the sixth term is: $a_{6} = 2^{6 - 1} = 2^{5}$ , so the product of the six first terms is: $P_{6} = \sqrt{(a_{1} \cdot a_{n})^{6}} = \sqrt{(1 \cdot 2^{5})^{6}} = \sqrt{2^{3} 0} = 2^{15} = 32.768$

To make the composition easier and to simplify the notation, if we are working with a large amount of numbers that we cannot write explicitly, to denote "the product of" we will use the Greek capital letter Pi: $\prod .$

We will write the variable we are multiplying and the initial term on the bottom, and the last term to be multiplied on the top. Next to the letter Pi, we will write the general term of the progression we want to multiply.

In the previous example, we will multiply the first six powers of two with: $P_{6} = \prod_{n = 1}^{6} 2^{n - 1}$

And multiplying the first three hundred terms of the succession $a_{n} = - 2 (- \frac{3}{7})^{n}$ we will write it as: $P_{300} = \prod_{n = 1}^{300} - 2 (- \frac{3}{7})^{n}$