tough problems of Arithmetic progression
Answers
Answer:
{\displaystyle \ a_{n}=a_{1}+(n-1)d},
and in general
{\displaystyle \ a_{n}=a_{m}+(n-m)d}{\displaystyle \ a_{n}=a_{m}+(n-m)d}.
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.
Step-by-step explanation:
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
{\displaystyle a_{1}a_{2}\cdots a_{n}=d{\frac {a_{1}}{d}}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)=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}=d^{n}{\frac {\Gamma \left(a_{1}/d+n\right)}{\Gamma \left(a_{1}/d\right)}},}{\displaystyle a_{1}a_{2}\cdots a_{n}=d{\frac {a_{1}}{d}}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)=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}=d^{n}{\frac {\Gamma \left(a_{1}/d+n\right)}{\Gamma \left(a_{1}/d\right)}},}
where {\displaystyle x^{\overline {n}}}x^{\overline{n}} denotes the rising factorial and {\displaystyle \Gamma }\Gamma denotes the Gamma function. (The formula is not valid when {\displaystyle a_{1}/d}a_1/d is a negative integer or zero.)[citation needed]
This is a generalization from the fact that the product of the progression {\displaystyle 1\times 2\times \cdots \times n}1 \times 2 \times \cdots \times n is given by the factorial {\displaystyle n!}n! and that the product
{\displaystyle m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n}{\displaystyle m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n}
for positive integers {\displaystyle m}m and {\displaystyle n}n is given by
{\displaystyle {\frac {n!}{(m-1)!}}.}\frac{n!}{(m-1)!}.
Taking the example 3, 8, 13, 18, 23, 28, ..., the product of the terms of the arithmetic progression given by an = 3 + (n-1)×5 up to the 50th term is
{\displaystyle P_{50}=5^{50}\cdot {\frac {\Gamma \left(3/5+50\right)}{\Gamma \left(3/5\right)}}\approx 3.78438\times 10^{98}.}P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}.