# need help on math question?

determine the sum of the arithmetic series

9 + 21 + 33 +... + 693

thanks in advance..

### 4 Answers

- Anonymous8 years agoFavorite Answer
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

Sum=(a1+an)*n/2, where a1-first member of an arithmetic progression, an - the last member of an arithmetic progression, n - number of members of an arithmetic progression

an=a1+(n-1)d, where d - the difference between the members of an arithmetic progression

In your task a1=9, d=a2-a1=21-9=12, an=693

Therefore n=(an-a1)/d+1=(693-9)/12+1=58

Then Sum=(9+693)*58/2=20358

Answer: Sum=20358

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- AmorcitoLv 58 years ago
common difference, d= 12.

last term, an= a + (n-1)*d

693 = 9 + (n-1)*12 => n= 58.

sum, S= (n/2)* [ (a+an)/2 ]

=> S= (58/2) * [ (9 + 693)/ 2 ]

calculate :-)

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- Wayne DeguManLv 78 years ago
a = 9 and d = 12

=> nth term is a + (n - 1)d = 9 + 12(n - 1) = 12n - 3

693 = 12n - 3

=> 12n = 696

i.e. n = 696/12 = 58.....i.e. 58 terms

=> Sn = (n/2)[2a + (n - 1)d]

i.e. S58 = (58/2)[18 + (57)(12)] = 20,358

:)>

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- 8 years ago
add each number with 12 and then the answer again with 12..

but this will be right only if the last number is 633..

i thing you have made mistake in numbers

9+21+33+45+57+69+.........+633

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