My Math Forum Some formulas of Arithmetic progression/series

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 July 28th, 2014, 10:58 PM #1 Member   Joined: Jun 2014 From: Brighton Posts: 49 Thanks: 2 Some formulas of Arithmetic progression/series An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Example: 2,4,6,8,10….. Arithmetic Series : The sum of the numbers in a finite arithmetic progression is called as Arithmetic series. Example: 2+4+6+8+10….. nth term in the finite arithmetic series Suppose Arithmetic Series a1+a2+a3+…..an Then nth term an=a1+(n-1)d Where a1- First number of the series an- Nth Term of the series n- Total number of terms in the series d- Difference between two successive numbers Sum of the total numbers of the arithmetic series Sn=n/2*(2*a1+(n-1)*d) Where Sn – Sum of the total numbers of the series a1- First number of the series n- Total number of terms in the series d- Difference between two successive numbers Example: Find n and sum of the numbers in the following series 3 + 6 + 9 + 12 + x? Here a1=3, d=6-3=3, n=5 x= a1+(n-1)d = 3+(5-1)3 = 15 Sn=n/2*(2a1+(n-1)*d) Sn=5/2*(2*3+(5-1)3)=5/2*18 = 45 I hope the above formulae are helpful to solve your math problems. Last edited by skipjack; July 29th, 2014 at 03:48 AM.
 July 29th, 2014, 06:44 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms In a related note, you may find the gp command sumformal() useful. Let's say you want to find the sum of $3n^2-2n$. Just type Code: sumformal(3*n^2-2*n) and you get Code: n^3 + 1/2*n^2 - 1/2*n in which you can substitute whatever values you need (either by hand or with gp's subst() command).
August 6th, 2014, 03:29 AM   #3
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Quote:
 Originally Posted by CRGreathouse In a related note, you may find the gp command sumformal() useful. Let's say you want to find the sum of $3n^2-2n$. Just type Code: sumformal(3*n^2-2*n) and you get Code: n^3 + 1/2*n^2 - 1/2*n in which you can substitute whatever values you need (either by hand or with gp's subst() command).

 August 6th, 2014, 09:58 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 As the nth number is a1 + (n - 1)d, Sn = n(a1 + an)/2. Note that (a1 + an)/2 is the median of the n numbers, and is the "middle" number of the series if n is odd. It's slightly confusing to start by referring to a sequence of numbers and then keep switching between using the words "number" and "term".

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# sn=n/2 2a1 (n-1)d

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