How can we find the product of $n$ terms of an arithmetic progression where common difference is not unity?
I just want to know the last $3$ digits of $7 \times 23 \times 39 \times \ldots \times 2071$ where common difference is $16$.
[Math] How to find the product of n terms of an arithmetic progression where common difference is not unity
sequences-and-series
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Best Answer
There are two (unrelated) questions here. An answer to the first question is $$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$ To answer the second question, consider $n=\prod\limits_{k=0}^{129}(7+16k)$. Note that $7+16k=0\pmod{5}$ for every $k=3\pmod{5}$. Using this for $k=3$, $k=8$ and $k=13$ yields $n=0\pmod{125}$, that is, $n=125m$ for some integer $m$.
Likewise, $7+16k=-1\pmod{8}$ and the number of terms in the product which defines $n$ is $130$, which is even, hence $n=1\pmod{8}$. And $125=5\pmod{8}$ hence $n=5m\pmod{8}$, which implies $m=5n\pmod{8}$, that is, $m=5\pmod{8}$.
Finally, $n=125\cdot5\pmod{125\cdot8}$, that is, $n=\underline{\ \ \ \ \ \ \ \ }\pmod{1000}$.