We are given that there
will be (1/2) a litre after the first pouring, so considering two successive
pourings (n and (n+1)) with 1/2 litre in each before the nth pouring occurs:
1/2 × (1/n) = 1/(2n)
1/2 - 1/(2n) = (n-1)/2n
1/2 + 1/(2n) = (n+1)/2n
(n-1)/2n and (n+1)/2n in
each urn after the nth pouring
Then now consider the
(n+1)th pouring
(n+1)/2n × 1/(n+1) =
1/(2n)
(n+1)/(2n) - 1/(2n) =
n/(2n) = 1/2
Therefore this means that after
an odd number of pouring, there will be 1/2 a litre in each urn
Since 1997 is an odd
number, then there will be 1/2 a litre of water in each urn.
Answer:
<span>1/2</span>
Answer:
0.1590909999999001
Step-by-step explanation:
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Answer:
10 + 7 + 4 + . . .; n = 5
The series is arithmetic
Common difference between terms = (-3)
10 - 3 = 7
7 - 3 = 4
Or:
10 + (-3) = 7
7 + (-3) = 4
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<u>Another example</u>:
Arithmetic:
5, 9, 13, 17
common difference = 4
5 + 4 = 9
9 + 4 = 13
Geometric:
4, 8, 16, 32
4 × 2 = 8
8 × 2 = 16
16 × 2 = 32
common ratio = 2
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10 + 7 + 4 + . . .; n = 5
a₁ = 10
a₂ = 7
a₄ = a₃ + d = 4 + (-3) = 4 - 3 = 1 (d = common difference)
a₅ = a₄ - 3 = 1 - 3 = -2
Or with formula:

d = a₂ - a₁ = 7 - 10 = -3
aₙ = a₁ + (n - 1)d
fourth term = a₄ = 10 + (4 - 1)(-3) = 10 + 3(-3) = 10 + (-9) = 10 - 9 = 1
fifth term = a₅ = 10 + (5 - 1)(-3) = 10 + 4(-3) = 10 + (-12) = 10 - 12 = -2
Sum of the finite series (if n = 5):

Manual count: S₅ = 10 + 7 + 4 + 1 + (-2) = 20