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.
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!
50
374811518532311154222000151220025110120126973463765827396654565868436844654656654524547654453646683765476756646754565446544665485658447
21
To divide the 100 loaves of bread among 5 men in an arithmetic progression, we need to find the common difference, as it will determine the distribution of bread among them. Let's denote the common difference as "d".
Let's consider the amounts of bread each man receives:
First man: x loaves
Second man: x + d loaves
Third man: x + 2d loaves
Fourth man: x + 3d loaves
Fifth man: x + 4d loaves
According to the given conditions, the sum of the three largest shares is seven times the sum of the two smallest shares. Mathematically, this can be represented as:
(x + 2d) + (x + 3d) + (x + 4d) = 7[(x) + (x + d)]
Simplifying the equation, we have:
3x + 9d = 14x + 7d
6d = 11x
We need to find values of x and d that satisfy this equation and the condition that the total number of loaves distributed is 100.
Since finding the exact solution requires trial and error, let's use the substitution method to find one possible solution.
Let's assume x = 20 and substitute it into the equation:
6d = 11x
6d = 11 * 20
6d = 220
d = 220 / 6
d = 36.67 (approx.)
Now, let's calculate the distribution of bread for each man:
First man: x = 20 loaves
Second man: x + d = 20 + 36.67 ≈ 56.67 loaves
Third man: x + 2d = 20 + 2 * 36.67 ≈ 93.33 loaves
Fourth man: x + 3d = 20 + 3 * 36.67 ≈ 130 loaves
Fifth man: x + 4d = 20 + 4 * 36.67 ≈ 166.67 loaves
Since we cannot distribute fractions of loaves, we need to adjust the numbers slightly:
- Reduce the second man's share by rounding down to keep the total allocation at 100 loaves.
- Reduce the fifth man's share by rounding up to keep the total allocation at 100 loaves.
The final distribution would be:
First man: 20 loaves
Second man: 56 loaves (rounded down from 56.67)
Third man: 93 loaves
Fourth man: 130 loaves
Fifth man: 101 loaves (rounded up from 100.67)
Note: This solution is one of the possible solutions that satisfies the given conditions. To find all the possible solutions, you'd need to explore different values of x and d that satisfy the equation 6d = 11x and distribute the loaves accordingly.