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15 Jul 2020, 02:24
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C
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74% (02:15) correct
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A concert hall has seats arranged in 50 consecutive rows. The first row has 300 seats. and each successive row has 5 fewer seats. How many seats are in the entire hall?
A. 7.500
B. 8,625
C. 8,875
D. 17.250
E. 17.750
A. 7.500
B. 8,625
C. 8,875
D. 17.250
E. 17.750
15 Jul 2020, 02:29
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Bunuel
Seats in 1st row = 300
Seats in 2nd row = 300-5 = 295
Seats in 3rd row = 295-5 = 290
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SUm of such 50 terms = 300+295+290+....
First term, a = 300
COmmon Difference = 295-300 = 290-295 = -5
CONCEPT: Sum of n terms in an Arithmetic Progression, \(S_n = (\frac{n}{2})*[2a+(n-1)*d]\)
i.e. \(S_{50} = (\frac{50}{2})*[2*300+(50-1)*(-5)] = 25*(600-245) = 8875\)
Answer: Option C
15 Jul 2020, 03:07
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Here, we can use the equation
S = \(\frac{n}{2}\) (2a+(n-1)d)
n = 50
a = 300
d = -5
s = \(\frac{50}{2}\) (2*300 +(49)(-5))
s = 25 (600 - 245)
s = 25 * 355
s = 8875
Answer C
S = \(\frac{n}{2}\) (2a+(n-1)d)
n = 50
a = 300
d = -5
s = \(\frac{50}{2}\) (2*300 +(49)(-5))
s = 25 (600 - 245)
s = 25 * 355
s = 8875
Answer C
