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

= 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