# 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

### pseudoprimes in k(i)

341 is a pseudoprime to base 2 (which is in k(1). It is also a pseudoprime to base (341 + i). Hence it is a pseudoprime in k(i) also. Similarly 561, a Carmichael number is a pseudoprime in k(i) as it is pseudo to the base (187 + i) as well as the base (561 + i). Interestingly it is also pseudo to the base (561 + 2i).

### pseudoprimes in k(i) (contd)

Although 561 is a Carmichael number in k(1) it is only a pseudoprime in k(i). If we have a composite number consisting of two primes of form 4m+3 and if it happens to be a pseudo to a base, say 2, it is also pseudo to the base (number + i). Example: 341 = 11*31; this is pseudo to base 2. It is also pseudo to base (341 + i) and base ( 341 + 2i).

### failure functions - another example

Let our definition of a failure be a non-Carmichael number. Note 561, a Carmichael number, can be represented by the quadratic polynomial x^2 + x +55 where x = 22. Hence x^2 + x + 55 is the parent function. The relevant failure function is x = 29 + 25k; here k belongs to N. When the values of x generated by the failure function are substituted in the parent function we get only failures i.e. non-Carmichael numbers.

### pseudoprimes in k(i) (contd)

561 is a pseudoprime to any base of shape (11*k+I) where k belongs to N. This is in addition to the other bases indicated in the previous message.

### Kahun Papyrus, a 1800 BCE text, and arithmetic progressions

The KP http://planetmath.org/kahunpapyrusandarithmeticprogressions method was known to Ahmes in 1650 BCE

### Egyptian Middle Kingdom arithmetic progressions

In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book on the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and the same class of arithmetic proportions used in the RMP. The KP and RMP report the same arithmetic proportion method to find the largest term. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 meant 100/10 = 10 in the KP). Finally add column 11’s result, 3 3/4, to 10, and the largest term, 13 3/4.

In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.

Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.

The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.

C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: ”the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms”.

1. number of terms: 10

2. arithmetical progression difference: 1/8

3. arithmetic progression sum: 10

The scribe used the following facts to find the largest term.

1. one-half of differences, 1/16, times number of terms minus one, 9,

1/16 times 9 = 9/16

2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.

That is, the KP scribe used formula 1.0:1111111

(1/2)d(n-1) + S/n = Xn (formula 1.0)

with,

d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.

When n was odd, x (n/2) = S/n,

and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = … = x(n/2) = S/n,

Note that Robins-Shute omitted the sum divided by the number of terms (S/n):

A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = …) by using an identical arithmetic progression rule.