Friday

November 27, 2015
Posted by **M.A.** on Saturday, June 13, 2009 at 7:44am.

57, 59, 61, 63,...1233?

My friend calculated 588.

I am too concrete in my thinking and did it the hard way, listing all of the numbers on an Excel sheet and found that there were 619 numbers. Who is correct?

And how is this calculated?

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**Reiny**, Saturday, June 13, 2009 at 8:29amYou have an "arithmetice sequence"

with the first number as 57 and a common difference of 2

the general term of any arithmetic sequence is given by

term(n) = a + (n-1)d, were n is the term number,

a is the first term, and

d is the common difference

so we have 1233 = 57 + (n-1)(2)

1233 = 57 + 2n - 2

1178 = 2n

n = 589

so there are 589 terms

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**M.A.**, Saturday, June 13, 2009 at 8:48amI have actually printed out all of the numbers "57, 59, 61,63, all the way to 1233". There are 619 numbers. I double checked for errors that all of the numbers were correct. There are actually 619 numbers, for real. The mathematical formula you used does not match with the actual print-out of the real numbers. Please explain.

Can you run the real numbers as a double-check for your formula and see what shows up on a list of counted numbers? The formula does not appear to agree with reality.

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**Reiny**, Saturday, June 13, 2009 at 9:11amsorry, this is a well-known formula and is correct.

There is no need to "run the the real numbers as a double-check"

let's check some of the numbers

e.g.

term(4) = 57 + (4-1)(2)

= 57 + 6 = 63 ...check!

term(612) = 57 + (612-1)(2) = 1278

This was you answer, but obviously doesn't verify

my answer:

term(589) = 57 + (589-1)(2)

= 57 + 1176

= 1233 ... check!!

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**Reiny**, Saturday, June 13, 2009 at 9:18amhere is a quick way to verify my answer

and to show your list has an error in it.

go to Excel

enter 57 into A1

in A2 enter the formula "= A1 + 2"

highlight columns A down to your 612

and use the 'copy down' to enter the formula in all those cells

now look in A589

Well, well....

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**Reiny**, Saturday, June 13, 2009 at 9:44amJust noticed that 'Writeacher' already gave you a detailed series of steps how to do this in Excel.

http://www.jiskha.com/display.cgi?id=1244858200

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**M.A.**, Saturday, June 13, 2009 at 11:14amIn doing the Excel list, my column A589 has the number 1177 listed. In checking the list of real numbers again, I have found another major error. Thank you so much for your patient help in working with this mathematically challenged individual.But at least I am learning!

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**Reiny**, Saturday, June 13, 2009 at 11:42amcheck that in your Excel list such values as A1 = 57 ,A2 = 59 etc are correct

if A4 = 63 but A589 is not equal to 1233 there has to be something wrong with your cell formula, or God forbid, your EXcel program

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**farah**, Monday, February 21, 2011 at 9:34am25+10+15+15=