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A clock strikes once at 1:00, twice at 2:00, three times at 3:00 and so forth. How many times does it strike over a 12 hour period?
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so we are finding the sum?
1+2+3+4+5+6+7+8+9+10+11+12?
Alright we can set that up like 1,2,3,4,5...12
You can do that too but that's not the way of answering it
DaWizjr math 20-1?
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Since we need the sum of those numbers, the common difference here is 1, \[\large S_n=\frac{n}{2}[2a_1+(n-1)d] \\ \\ \large S_{12}=\frac{12}{2}[2(1)+(12-1)1]=?\]
Try that out :)
N=12 a1 or t1=1:00 d=?
d is 1, because 2-1=1
ok
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12=12/2[2(1)+(12-1)1]
guessing t1=1?
noo \(S_{12}\) is just a statement, \[\frac{12}{2}[2(1)+(12-1)1]=?\]
Yea
\(T_1=a_1=1\)
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I got 78
Correct
Nice
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