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#1
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A lecture hall has 40rows of seats. there are 10 seats in the the first row (a1=10) 12 seats in the 2nd row and so on, with two more seats in each row than in the previous row. How many seats are in the lecture hall?
Whats the equation and answer? |
#2
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88*19.5?
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#3
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10 seats in the first row. 2 more seats every row. so (39*2)+10 seats in the final row, or 88 if you like.
1st and 40th row have 98 combined seats 2nd and 39th row have same number of combined seats as 2nd row has 2 more than 1st, 39th row 2 less than 40. 20*98 is therefore the answer, ot 1960 |
#4
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(40*10)+((39*2)+(38*2)+(37*2)+(36*2)+(35*2)+...+2)
That's how I roll. Man, I would hate to do this with 400 rows! <font color="white">This is why I got a C in Calc II</font> |
#5
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Uh, 80*19.5 + 400 = 1960?
Edit: 78*20 + 400 = 1960 would make more sense I guess, but that's not how my brain did it for some reason. GoT |
#6
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Wouldnt it be 98 * 20, since in every 2 row extreme opposite combo, there are 98 total seats (like the first row is 10, last row is 88, so 98 total seats)?
And what is the equation for this? |
#7
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pi r^2 radius. or maybe it was pi r^2 friction
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#8
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[ QUOTE ]
And what is the equation for this? [/ QUOTE ] I'm probably wrong here, but I'd guess it looks something like 39 [Sum]10 + 2n n=0 |
#9
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Just put it in excel, drag the column down to how many rows there are, sum it up. Who needs math.
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#10
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let i= counting device.
sum from i=0 to 39 of (10+2i) or sum from i=1 to 40 of (8+2i) |
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