Solution:
ARITHMETIC SEQUENCE:
Finding the 4 arithmetic means between 68 and 3 thus we have 6th term all in all:
Let a₂, a₃, a₄ and a₅ be the 4 arithmetic means missing:
So we apply the given values to get the missing terms:
First, we get the common difference of the terms,
That is,
a₆ = a₁ + (6-1)d
3 = 68 + (6-1)d
3 = 68 + 5d
3-68 = 5d
-65/5 = 5d/5
d = -13
So, we substitute the value of common difference and the known values,
a₂ = a₁ + (n-1)d
a₂ = 68 + (1)(-13)
a₂ = 68+(-13)
a₂ = 55
a₃ = a₁ + (n-1)d
a₃ = 68 + (3-1)(-13)
a₃ = 68 + (2)(-13)
a₃ = 68+(-26)
a₃ = 42
a₄ = a₁ + (n-1)d
a₄ = 68 + (4-1)(-13)
a₄ = 68 + (3)(-13)
a₄ = 68+(-39)
a₄ = 29
a₅ = a₁ + (n-1)d
a₅ = 68 + (5-1)(-13)
a₅ = 68 + (4)(-13)
a₅ = 68+(-52)
a₅ = 16
Therefore, the sequence are 68, 55, 42, 29, 16, 3.
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Answers & Comments
Solution:
ARITHMETIC SEQUENCE:
Finding the 4 arithmetic means between 68 and 3 thus we have 6th term all in all:
Let a₂, a₃, a₄ and a₅ be the 4 arithmetic means missing:
So we apply the given values to get the missing terms:
First, we get the common difference of the terms,
That is,
a₆ = a₁ + (6-1)d
3 = 68 + (6-1)d
3 = 68 + 5d
3-68 = 5d
-65/5 = 5d/5
d = -13
So, we substitute the value of common difference and the known values,
a₂ = a₁ + (n-1)d
a₂ = 68 + (1)(-13)
a₂ = 68+(-13)
a₂ = 55
a₃ = a₁ + (n-1)d
a₃ = 68 + (3-1)(-13)
a₃ = 68 + (2)(-13)
a₃ = 68+(-26)
a₃ = 42
a₄ = a₁ + (n-1)d
a₄ = 68 + (4-1)(-13)
a₄ = 68 + (3)(-13)
a₄ = 68+(-39)
a₄ = 29
a₅ = a₁ + (n-1)d
a₅ = 68 + (5-1)(-13)
a₅ = 68 + (4)(-13)
a₅ = 68+(-52)
a₅ = 16
Therefore, the sequence are 68, 55, 42, 29, 16, 3.