The
product of the members of a finite arithmetic progression with an initial element
a1, common differences
d, and
n elements in total is determined in a closed expression :\begin{align} a_1 a_2 a_3 \cdots a_n &= a_1 (a_1+d) (a_1+2d) \cdots (a_1+(n-1)d) \\[1ex] &= \prod_{k=0}^{n-1} (a_1+kd) = d^n \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}} \end{align} where \Gamma denotes the
Gamma function. The formula is not valid when a_1/d is negative or zero. This is a generalization of the facts that the product of the progression 1 \times 2 \times \cdots \times n is given by the
factorial n! and that the product :m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n for
positive integers m and n is given by :\frac{n!}{(m-1)!}.
Derivation :\begin{align} a_1a_2a_3\cdots a_n &=\prod_{k=0}^{n-1} (a_1+kd) \\[2pt] &= \prod_{k=0}^{n-1} d\left(\frac{a_1}{d}+k\right) \\[2pt] &= d \left (\frac{a_1}{d}\right) d \left (\frac{a_1}{d}+1 \right )d \left ( \frac{a_1}{d}+2 \right )\cdots d \left ( \frac{a_1}{d}+(n-1) \right ) \\[2pt] &= d^n\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right)=d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}} \end{align} where x^{\overline{n}} denotes the
rising factorial. By the recurrence formula \Gamma(z+1)=z\Gamma(z), valid for a
complex number z>0, :\Gamma(z+2)=(z+1)\Gamma(z+1)=(z+1)z\Gamma(z), :\Gamma(z+3)=(z+2)\Gamma(z+2)=(z+2)(z+1)z\Gamma(z), so that : \frac{\Gamma(z+m)}{\Gamma(z)} = \prod_{k=0}^{m-1}(z+k) for m a positive integer and z a positive complex number. Thus, if a_1/d > 0 , :\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right)= \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}}, and, finally, :a_1a_2a_3\cdots a_n = d^n\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right) = d^n \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}}
Examples ;Example 1 Taking the example 3, 8, 13, 18, 23, 28, \ldots , the product of the terms of the arithmetic progression given by a_n = 3 + 5(n-1) up to the 50th term is :P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}. ; Example 2 The product of the first 10 odd numbers (1,3,5,7,9,11,13,15,17,19) is given by : 1\cdot 3\cdot 5\cdots 19 =\prod_{k=0}^{9} (1+2k) = 2^{10} \cdot \frac{\Gamma \left(\frac{1}{2} + 10\right) }{\Gamma \left( \frac{1}{2} \right) } = ==Standard deviation==