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Tom is a Door

[Math Help] Arithmatic Sequence

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Posted · Original PosterOP

Given: summation of A1 through A10 is 235; and summation of A11 through A20 is 735. Problem asks to find A1 and the difference in this sequence
Thank You

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I feel like this should be in programming. /s


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S = n(2a+(n-1)d)/2

S = sum

n = number of digits to sum (10)

a = first number in the series (A1)

d = difference between numbers (1)

 

 

 

For the first part: 
19->28
235 = 10(2a+(10-1)1/2
235 = 10(2a+9)/2
235 = 5(2a+9)
47 = 2a+9
38 = 2a
19 = a

So A1 = 19.

 

Part two should be pretty easy to figure out now.

 


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26 minutes ago, VerticalDiscussions said:

maxresdefault.jpg (1920×1080)

It's the generic formula behind S = n(n+1)/2, that you've probably seen/heard at some point used to calculate the sum from 1 to 100 (100*101/2) = 5050.


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21 minutes ago, djdwosk97 said:

It's the generic formula behind S = n(n+1)/2, that you've probably seen/heard at some point used to calculate the sum from 1 to 100 (100*101/2) = 5050.

 

maxresdefault.jpg


- snip-

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3 minutes ago, MrDynamicMan said:

 

-

Edumacate yourself. 


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19 hours ago, Tom is a Door said:

Given: summation of A1 through A10 is 235; and summation of A11 through A20 is 735. Problem asks to find A1 and the difference in this sequence
Thank You

For this I took that to get S10 it is 10a+45r=235 (a being the first term, r being the difference between each term) 

Thus S11 to S20 is 10a+145r=735 (as it is basically the same however offset by 10 positions or Rs, this then adds up to an additional 100r)

take the two away fro each other and you are left with 100r=500 thus r=5

(then plug it back into one of the equations above) 10a+45x5=235    you get 10a=10 so a=1

at this point you have your answers, however i plugged it back into the general formula to check I was right.

(n(a1+an))/2  (10(1+46))/2 which does then equal the 235

 

Thanks for a nice problem, hope that helped you out.

 

thing

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