Asked by jake
divide 100 loaves of bread amoung 5 men so that the amounts are in arithmetic progression ( increase by the same amount each time) and the sum of the three largest shares is seven times the sum of the two smallest.
Answers
Answered by
Reiny
use the formula for the general term of an arithmetic sequence
the terms would be
a, a+d, a+2d, a+3d, and a+4d
adding them up: 5a + 10d = 100
or a + 2d = 20 after dividing by 5
given:
sum of 3 largest equals seven times the sum of the two smallest, or
a+2d + a+3d + a+4d = 7(a + a+d)
3a + 9d = 14a + 7d
2d = 11a
from first equation a = 20 - 2d
into the second:
2d = 11(20-2d)
2d = 220 - 22d
24d = 220
d = 220/24 = 55/6
subbing back: a = 20 - 2(55/6) = 5/3
checking:
first term is 5/3 or 10/6
is 10/6 + 65/6 + 120/6 + 175/6 + 230/6 equal to 100 ?
YES!
the terms would be
a, a+d, a+2d, a+3d, and a+4d
adding them up: 5a + 10d = 100
or a + 2d = 20 after dividing by 5
given:
sum of 3 largest equals seven times the sum of the two smallest, or
a+2d + a+3d + a+4d = 7(a + a+d)
3a + 9d = 14a + 7d
2d = 11a
from first equation a = 20 - 2d
into the second:
2d = 11(20-2d)
2d = 220 - 22d
24d = 220
d = 220/24 = 55/6
subbing back: a = 20 - 2(55/6) = 5/3
checking:
first term is 5/3 or 10/6
is 10/6 + 65/6 + 120/6 + 175/6 + 230/6 equal to 100 ?
YES!
Answered by
Anonymous
50
Answered by
bich
374811518532311154222000151220025110120126973463765827396654565868436844654656654524547654453646683765476756646754565446544665485658447
Answered by
Kaptain Mexican
21
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