## New Articles

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*Rec*]**Kenosymplirostic numbers**by imaginary.iAug 14[

*Edu*]**How to find whether a given number is prime or not...**by burgessAug 12[

*Edu*]**BODMAS Rule application**by burgessAug 8[

*Edu*]**Tests of Divisibility- Simple tricks**by burgessAug 7[

*Res*]**0/0 is possible and has an answer**by imaginary.iAug 2[

*Ref*]**Sophomore's dream**by pahioJul 9[

*Res*]**examples of growth of perturbations in chemical or...**by rspuzioMay 24[

*Ref*]**proof of Stirling's approximation**by rspuzioMay 8[

*Res*]**Example of stochastic matrix of mapping**by rspuzioApr 23[

*Res*]**6. Discussion**by rspuzioApr 20[

*Res*]**5. Entanglement**by rspuzioApr 20[

*Res*]**4. Measurement**by rspuzioApr 20[

*Res*]**3. Distributed dynamical systems**by rspuzioApr 20[

*Res*]**2. Stochastic maps**by rspuzioApr 20## Latest Messages

Aug 18

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Aug 9

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Aug 7

Aug 6

Aug 4

Aug 2

Aug 1

Jul 31

Jul 29

Jul 29

\begin{itemize}
\item
\end{itemize}\begin{flushleft}
\end{flushleft}
\textbf{}gg

Aug 17

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.

Aug 14

There is an exception to "years divisible by 4 being leap years": If a
year is divisible by 100 (such years are divisible by 4), such a year is
a leap year ONLY IF it is also divisible by 400. For example, 2000 was a
leap year, but 1900 was not. Shockley's "Introduction to Number Theory"
contains a "day of the week" formula that includes the above fact.

Aug 9

Carmichael numbers constitute a sub-set of Devaraj numbers. To understand more about them
refer sequences A 104016, A 104017 and A 162290. Some interesting facts pertaining to
them will follow.

Aug 8

The reason was a dollar-sign error =o)
Corrected!
BTW, the PM search engine has long been out of order. It's quite
difficult to find entries on a wanted subject.

Aug 7

Pl. see the older entry http://planetmath.org/divisibilitytest

Aug 7

Definition: These are the impossible prime factors of 3^n - 2 (n belongs to N).
This is identical with the sequence A123239 (OEIS ).

Aug 6

Nice solution

Aug 4

Please see e.g. the entry http://planetmath.org/division
BTW, one cannot write "P(R)*0 = 0" since your P(R) is not a _number_

Aug 2

under repair ... For now consider a \PMlinkexternal{generalized scribal square root method}{http://planetmath.org/squarerootof3567and29} was published in Dec. 2012.
I INTRODUCTION: An ancient square root method was decoded in 2012. Scholars for 100 years failed to fully decode Archimedes’ three steps that estimated unit fraction series answers to four to six places (modern standards), a method used by Fibonacci and Galileo.
Unresolved aspects of ancient ESTIMATION OF PI problem were reported by Kevin Brown and E.B. Davis with upper and lower limits;
(1351/780)^2 is greater than PI is greater than (265/153)^2
A. Archimedes’ actual square root of pi^2 and n^2 method, decoded in Dec 2012 and Jan 2013, calculated the higher PI limit(1351/780)^2 by:
1. step 1. guess (1 + 2/3)^2 = 1 + 4/3 + 4/9 = 2 + 3/9 + 4/9, meant 2/9 = error1
2. step 2 reduced error1 2/9
by dividing 2/9 by 2(1 + 2/3)
steps that meant
2/9 x (3/10) = 1/15
such that
(1 + 2/3 + 1/15)^2, error2 (1/15)^2 = 1/225 = error2
knowing (1 + 11/15) = 26/15
3. step 3 reduced error2 = 1/225 by dividing by 2 x (26/15) = 52/15
1/225 x (15/52) = 1/15 x (1/52) = (1/780)^2 = error 3
reached
(26/15 - 1/780)^2 = (1351/780)^2 in modern fractions
recorded a unit fraction series that began with step 2 data and subtracted 1/780
(1 + 2/3 + 1/15 - 1/780)^2
as Archimedes would have written
(1 + 2/3 + 1/30 + (13 + 6 + 4 + 2)/780)^2 = (1 + 2/3 + 1/30 + 1/60 + 1/+ 1/195 + 1/390)^2
B . The lower limit 265/153 modified step 2, used
1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51)
such that (1 + 111/153)changed to (1 + 112/153) = 265/153
II. PROOF: Modern translations of scribal square roots of five (5), six (6), seven (7), (29) and any rational numberPlanetmathPlanetmathPlanetmath are demonstrated below (A, B, C, D).
Greek and Egyptian algebraic steps were finite. Decoding algorithmic looking finite arithmetic steps 2, 3 and 4 have been demystified.
A. Computed the square root of five(5) that estimated
(Q + R)^2, R= 1/(1/2Q)
step 1: estimated
Q = 2. R,= (1/4) such that
(2 + 1/4)^2 = 5 + (1/4)^2; error1 = 1/16
step 2, reduced error1 that divided
1/16 by 2(2 + 1/4)= 18/4 such that
1/16 x (4/18) = 1/72 = error2 = (1/72)^2
2 + 1/4 - (1/72)^2)^2 = (2 + 1285/5184)^2;
as Archimedes and Ahmes would have re-written
(2 + 1285/5184) as a unit fraction series:
a [2 + (864 + 398 + 51 + 1/17 + 6)/5184) =
[2 + 1/6 + 1/13 + 1/39 + 1/864]^2
b [2 + (864 + 370 + 64 + 6 +1 )/5184] =
[2 + 1/6 + 1/14 + 1/81 + 1/864 + 1/5184]^2
Note that scribal shorthand notes suggested by academics prior to Dec. 2012 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been found in the medieval era. In May 2013 the shorthand notes of Galileo reveal the same method was also used by Fibonacci and Archimedes.
In the square root of five step 3 was not needed.
B. square root of six (6) ,
step 1: estimated Q = 2, R = (6 -4)/4 = 1/2 such that
(2 + 1/2)^2 = 6 + (1/2)^2, error1 = 1/4
step 2: reduced 1/4 error that divided by (2 + 1/2)
such that
1/4 x (2/10) = 1/20.
hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400
Ahmes, Archimes and Fibonacci may have stopped at this point and recorded
(2 + 99/400) as a unit fraction series
[2 + (80 + 10 + 8 + 1)/400] =
[2 + 1/5 + 1/40 + 1/50 + 1/400 ]
step 3 (as included Archimedes square root of three method) was optionasl
divided 1/400 by (400/1798) = 1/1798,
hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2
Archimedes’ actual square root method would have recorded
[2+ 1/5 + 1/40 + 1/50 + 1/400]
with a note that a longer series, with an error of (1/1798)^2 was easily found.
C. square root of seven (7)
step 1: estimates Q = 2, R = 3/4 and (2 + 3/4)^2 =
7 + (3/4)^2 , error1 = 9/16
step 2: divides 9/16 by twice (2 + 3/4) =
(9/16)(4/22) = 9/88 = (1/11 + 1/88)
(2 + 3/4 -9/88) = [2 + 1/2 + 13/88]^2 =
[2 + 1/2 + (8 + 4 + 1)/88]^2 =
[2 + 1/2 + 1/11 + 1/22 + 1/88]^2
step 3 may have been required
divide 9/88 by twice (2 + 1/2 + 13/88) =
(9/88)(88/466) =
9/466 = (1/155 + 1/155 + 1/466)
hence [2 + 1/2 + 1/11 + 1/22 + 1/88 - (2/155 +1/466)]^2 would have been recorded as a unit fraction series
Note that Archimedes and Ahmes paired
(1/22 - 2/155) =
and
(1/88 - 1/466) =
readers may choose the most likely final unit fraction series,
D. ESTIMATE the square root of 29.
step 1 found R = (29-25)/2Q = 4/10 = 2/5
such that
(5 + 2/5)^2 = 29 + 4/25 = error1
STEP 2
reduce error1 4/25 by dividing by 2(5 + 2/5)
4/25 x 5/54 = 2/27
hence a final unit fraction series converted
(5 + 2/5 - 2/27)^2 by considering
(5 + 1/5 + (27-10)/135)^2 = (5 + 1/5 + 1/9 + 2/135)^2
was accurate to (2/27)^2
NOTE: the conversion of 2/135 to a unit fraction series followed Ahmes 2/n table rules
(2/5)(1/27) = (1/3 + 1/15)(1/27) = 1/51 + 1/405
MEANT THE SQUARE ROOT OF 29 WAS ESTIMATED IN TWO STEPS BY
(5 + 1/5 + 1/9 + 1/51 + 1/405)^2
footnote: Fibonacci’s square root of 17 method was appropriately cited as used by Galileo though not properly analyzed in every detail. Fibonacci guessed (4 + 1/8)^2 = (17 + 1/64) , and Fibonacci reduced the estimated 1/64 error foumd an inversePlanetmathPlanetmathPlanetmath proportion:1/64 x 8/66 = 1/528 which meant (4 + 1/8 - 1/528)^2 = (2177/528)^2 = 17.000003 is accurate to (1/528)^2 perArchimedes and not by Newton, as suggested by scholars prior to 2012.
III CONCLUSION
Unit fractionPlanetmathPlanetmath square root was formalized by 2050 BCE and used by Egyptians, Greeks, Arabs, medieval scribes and as late as Galileo. The method estimated irrational square roots of N by 1-step, 2-step, 3-step and 4-steps methods. Step 1 guessed quotients (Q) and remainders (R) = n/(2Q) with n = (N - Q^2). Step 2, 3, and 4 reduced error 1, 2 and 3 associated with the previous step by dividing by 2(Q + R).
References
1 A.B. Chace, Bull, L, Manning, H.P., Archibald, R.C., The Rhind Mathematical Papyrus, Mathematical
2 Marshall Clagett Ancient Egyptian Science, Volume III, American Philosophical Society, Philadelphia, 1999.
3 Richard Gillings, Mathematics in the Time of the Pharaohs, Dover Books, 1992, PAGE 214-217.
4 H. Schack-Schackenburg, ”Der Berliner Papyreys 6619”, Zeitscrift fur Agypyische Sprache , Vol 38 (1900), pp. 135-140 and Vol. 40 (1902), p. 65f.
5 L. E. Sigler, Fibonacci’s Liber Abaci, Leonardo’s Book of Calculation ,Springer, NY, 2002, page 491.

Aug 1

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.

Jul 31

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).

Jul 29

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).

Jul 29

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>

## Latest Messages

Aug 18

Aug 17

Aug 14

Aug 9

Aug 8

Aug 7

Aug 7

Aug 6

Aug 4

Aug 2

Aug 1

Jul 31

Jul 29

Jul 29

\begin{itemize}
\item
\end{itemize}\begin{flushleft}
\end{flushleft}
\textbf{}gg

Aug 17

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.

Aug 14

There is an exception to "years divisible by 4 being leap years": If a
year is divisible by 100 (such years are divisible by 4), such a year is
a leap year ONLY IF it is also divisible by 400. For example, 2000 was a
leap year, but 1900 was not. Shockley's "Introduction to Number Theory"
contains a "day of the week" formula that includes the above fact.

Aug 9

Carmichael numbers constitute a sub-set of Devaraj numbers. To understand more about them
refer sequences A 104016, A 104017 and A 162290. Some interesting facts pertaining to
them will follow.

Aug 8

The reason was a dollar-sign error =o)
Corrected!
BTW, the PM search engine has long been out of order. It's quite
difficult to find entries on a wanted subject.

Aug 7

Pl. see the older entry http://planetmath.org/divisibilitytest

Aug 7

Definition: These are the impossible prime factors of 3^n - 2 (n belongs to N).
This is identical with the sequence A123239 (OEIS ).

Aug 6

Nice solution

Aug 4

Please see e.g. the entry http://planetmath.org/division
BTW, one cannot write "P(R)*0 = 0" since your P(R) is not a _number_

Aug 2

under repair ... For now consider a \PMlinkexternal{generalized scribal square root method}{http://planetmath.org/squarerootof3567and29} was published in Dec. 2012.
I INTRODUCTION: An ancient square root method was decoded in 2012. Scholars for 100 years failed to fully decode Archimedes’ three steps that estimated unit fraction series answers to four to six places (modern standards), a method used by Fibonacci and Galileo.
Unresolved aspects of ancient ESTIMATION OF PI problem were reported by Kevin Brown and E.B. Davis with upper and lower limits;
(1351/780)^2 is greater than PI is greater than (265/153)^2
A. Archimedes’ actual square root of pi^2 and n^2 method, decoded in Dec 2012 and Jan 2013, calculated the higher PI limit(1351/780)^2 by:
1. step 1. guess (1 + 2/3)^2 = 1 + 4/3 + 4/9 = 2 + 3/9 + 4/9, meant 2/9 = error1
2. step 2 reduced error1 2/9
by dividing 2/9 by 2(1 + 2/3)
steps that meant
2/9 x (3/10) = 1/15
such that
(1 + 2/3 + 1/15)^2, error2 (1/15)^2 = 1/225 = error2
knowing (1 + 11/15) = 26/15
3. step 3 reduced error2 = 1/225 by dividing by 2 x (26/15) = 52/15
1/225 x (15/52) = 1/15 x (1/52) = (1/780)^2 = error 3
reached
(26/15 - 1/780)^2 = (1351/780)^2 in modern fractions
recorded a unit fraction series that began with step 2 data and subtracted 1/780
(1 + 2/3 + 1/15 - 1/780)^2
as Archimedes would have written
(1 + 2/3 + 1/30 + (13 + 6 + 4 + 2)/780)^2 = (1 + 2/3 + 1/30 + 1/60 + 1/+ 1/195 + 1/390)^2
B . The lower limit 265/153 modified step 2, used
1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51)
such that (1 + 111/153)changed to (1 + 112/153) = 265/153
II. PROOF: Modern translations of scribal square roots of five (5), six (6), seven (7), (29) and any rational numberPlanetmathPlanetmathPlanetmath are demonstrated below (A, B, C, D).
Greek and Egyptian algebraic steps were finite. Decoding algorithmic looking finite arithmetic steps 2, 3 and 4 have been demystified.
A. Computed the square root of five(5) that estimated
(Q + R)^2, R= 1/(1/2Q)
step 1: estimated
Q = 2. R,= (1/4) such that
(2 + 1/4)^2 = 5 + (1/4)^2; error1 = 1/16
step 2, reduced error1 that divided
1/16 by 2(2 + 1/4)= 18/4 such that
1/16 x (4/18) = 1/72 = error2 = (1/72)^2
2 + 1/4 - (1/72)^2)^2 = (2 + 1285/5184)^2;
as Archimedes and Ahmes would have re-written
(2 + 1285/5184) as a unit fraction series:
a [2 + (864 + 398 + 51 + 1/17 + 6)/5184) =
[2 + 1/6 + 1/13 + 1/39 + 1/864]^2
b [2 + (864 + 370 + 64 + 6 +1 )/5184] =
[2 + 1/6 + 1/14 + 1/81 + 1/864 + 1/5184]^2
Note that scribal shorthand notes suggested by academics prior to Dec. 2012 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been found in the medieval era. In May 2013 the shorthand notes of Galileo reveal the same method was also used by Fibonacci and Archimedes.
In the square root of five step 3 was not needed.
B. square root of six (6) ,
step 1: estimated Q = 2, R = (6 -4)/4 = 1/2 such that
(2 + 1/2)^2 = 6 + (1/2)^2, error1 = 1/4
step 2: reduced 1/4 error that divided by (2 + 1/2)
such that
1/4 x (2/10) = 1/20.
hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400
Ahmes, Archimes and Fibonacci may have stopped at this point and recorded
(2 + 99/400) as a unit fraction series
[2 + (80 + 10 + 8 + 1)/400] =
[2 + 1/5 + 1/40 + 1/50 + 1/400 ]
step 3 (as included Archimedes square root of three method) was optionasl
divided 1/400 by (400/1798) = 1/1798,
hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2
Archimedes’ actual square root method would have recorded
[2+ 1/5 + 1/40 + 1/50 + 1/400]
with a note that a longer series, with an error of (1/1798)^2 was easily found.
C. square root of seven (7)
step 1: estimates Q = 2, R = 3/4 and (2 + 3/4)^2 =
7 + (3/4)^2 , error1 = 9/16
step 2: divides 9/16 by twice (2 + 3/4) =
(9/16)(4/22) = 9/88 = (1/11 + 1/88)
(2 + 3/4 -9/88) = [2 + 1/2 + 13/88]^2 =
[2 + 1/2 + (8 + 4 + 1)/88]^2 =
[2 + 1/2 + 1/11 + 1/22 + 1/88]^2
step 3 may have been required
divide 9/88 by twice (2 + 1/2 + 13/88) =
(9/88)(88/466) =
9/466 = (1/155 + 1/155 + 1/466)
hence [2 + 1/2 + 1/11 + 1/22 + 1/88 - (2/155 +1/466)]^2 would have been recorded as a unit fraction series
Note that Archimedes and Ahmes paired
(1/22 - 2/155) =
and
(1/88 - 1/466) =
readers may choose the most likely final unit fraction series,
D. ESTIMATE the square root of 29.
step 1 found R = (29-25)/2Q = 4/10 = 2/5
such that
(5 + 2/5)^2 = 29 + 4/25 = error1
STEP 2
reduce error1 4/25 by dividing by 2(5 + 2/5)
4/25 x 5/54 = 2/27
hence a final unit fraction series converted
(5 + 2/5 - 2/27)^2 by considering
(5 + 1/5 + (27-10)/135)^2 = (5 + 1/5 + 1/9 + 2/135)^2
was accurate to (2/27)^2
NOTE: the conversion of 2/135 to a unit fraction series followed Ahmes 2/n table rules
(2/5)(1/27) = (1/3 + 1/15)(1/27) = 1/51 + 1/405
MEANT THE SQUARE ROOT OF 29 WAS ESTIMATED IN TWO STEPS BY
(5 + 1/5 + 1/9 + 1/51 + 1/405)^2
footnote: Fibonacci’s square root of 17 method was appropriately cited as used by Galileo though not properly analyzed in every detail. Fibonacci guessed (4 + 1/8)^2 = (17 + 1/64) , and Fibonacci reduced the estimated 1/64 error foumd an inversePlanetmathPlanetmathPlanetmath proportion:1/64 x 8/66 = 1/528 which meant (4 + 1/8 - 1/528)^2 = (2177/528)^2 = 17.000003 is accurate to (1/528)^2 perArchimedes and not by Newton, as suggested by scholars prior to 2012.
III CONCLUSION
Unit fractionPlanetmathPlanetmath square root was formalized by 2050 BCE and used by Egyptians, Greeks, Arabs, medieval scribes and as late as Galileo. The method estimated irrational square roots of N by 1-step, 2-step, 3-step and 4-steps methods. Step 1 guessed quotients (Q) and remainders (R) = n/(2Q) with n = (N - Q^2). Step 2, 3, and 4 reduced error 1, 2 and 3 associated with the previous step by dividing by 2(Q + R).
References
1 A.B. Chace, Bull, L, Manning, H.P., Archibald, R.C., The Rhind Mathematical Papyrus, Mathematical
2 Marshall Clagett Ancient Egyptian Science, Volume III, American Philosophical Society, Philadelphia, 1999.
3 Richard Gillings, Mathematics in the Time of the Pharaohs, Dover Books, 1992, PAGE 214-217.
4 H. Schack-Schackenburg, ”Der Berliner Papyreys 6619”, Zeitscrift fur Agypyische Sprache , Vol 38 (1900), pp. 135-140 and Vol. 40 (1902), p. 65f.
5 L. E. Sigler, Fibonacci’s Liber Abaci, Leonardo’s Book of Calculation ,Springer, NY, 2002, page 491.

Aug 1

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.

Jul 31

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).

Jul 29

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).

Jul 29

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>