Wikipedia:Reference desk/Archives/Mathematics/2020 October 2

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October 2

how is the #5 and #6 formulae of Arithmetic progression been derived?

I wanted to know how is the #5 and #6 formulae of Arithmetic progression been derived. I know the derivation of $a_{n}$ or the nth term of an A.P. and the derivation of $S_{n}$ or the sum of nth terms of an A.P. — Preceding unsigned comment added by Huzaifa abedeen (talk • contribs) 10:04, 2 October 2020 (UTC)

The arithmetic mean of a bag of values is equal to their sum (given by #4) divided by the number of elements $n$. So divide the formula at #4 by $n$.

For an arithmetic progression starting at $a_{1}$ with increment $d$, we have:

$a_{1}=a_{1},~a_{2}=a_{1}+d,~a_{3}=a_{1}+2d,~a_{4}=a_{1}+3d,~...,~a_{n}=a_{1}+(n-1)d.$

The equality $a_{n}=a_{1}+(n-1)d$ can easily be formally proved by mathematical induction. Since $n$ is an arbitrary index, it is also the case that $a_{m}=a_{1}+(m-1)d$. Subtract these two equations from each other, and solve for $d$.  --Lambiam 17:10, 2 October 2020 (UTC)

Derivation of ${\overline {a}}$ or the arithmetic mean

$S_{n}={\frac {n}{2}}(a_{1}+a_{n})$

$\Rightarrow {\frac {S_{n}}{n}}={\frac {{\frac {n}{2}}(a_{1}+a_{n})}{n}}$

$\Rightarrow {\overline {a}}={\frac {a_{1}+a_{n}}{2}}$

Derivation of $d$ or the common difference

$a_{m}-a_{n}={a_{1}+(m-1)d}-{a_{1}+(n-1)d}$

$\Rightarrow a_{m}-a_{n}=(m-1)d-(n-1)d$

$\Rightarrow a_{m}-a_{n}=d\{{(m-1)-(n-1)}\}$

$\Rightarrow a_{m}-a_{n}=d\{m-n\}$

$\Rightarrow d={\frac {a_{m}-a_{n}}{m-n}}$

Thank you Lambian sir. You are a great mathematician. Huzaifa abedeen (talk) 09:34, 7 October 2020 (UTC) --Huzaifa abedeen