Testwiki: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 $\bar{a}$ or the arithmetic mean

$S_n=\frac{n}{2}(a_1+a_n)$

$\Rightarrow \frac{S_n}{n}=\frac{\frac{n}{2}(a_1+a_n)}{n}$

$\Rightarrow \bar{a}=\frac{a_1+a_n}{2}$

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