The (n+2) term of an arithmetic progression can be given by the formula:
[tex]\[ a_{n+2} = a + (n+1)d \][/tex]
Given:
[tex]a = -4[/tex]
[tex]l = 16 [/tex]
[tex]a_{n+2} = l = 16[/tex]
We can write:
[tex]16 = -4 + (n+1)d[/tex]
[tex]16 + 4 = (n+1)d[/tex]
[tex]20 = (n+1)d[/tex]
Case 1: Inserting 2 Arithmetic Means
lnsert (2) arithmetic means (n = 2):
[tex]20 = (2+1)d[/tex]
[tex]20 = 3d[/tex]
[tex]d = \frac{20}{3} = \frac{20}{3} \approx 6.67[/tex]
The sequence in this case is:
[tex]-4, -4 + \frac{20}{3}, -4 + 2 \left(\frac{20}{3}\right), 16[/tex]
Simplifying:
[tex]{-4, \frac{-4+20}{3} = \frac{16}{3} \approx 5.33, \frac{-4+40}{3} = \frac{36}{3} = 12, 16}[/tex]
Case 2: Inserting 3 Arithmetic Means
lnsert (3) arithmetic means (n = 3):
[tex]20 = (3+1)d[/tex]
[tex]20 = 4d[/tex]
[tex]d = \frac{20}{4} = 5[/tex]
The sequence in this case is:
[tex]{-4, -4 + 5, -4 + 2 \cdot 5, -4 + 3 \cdot 5, 16}[/tex]
Simplifying:
[tex]-4, 1, 6, 11, 16 [/tex]
General Formula
In general, if you want (n) arithmetic means, the common difference (d) is:
[tex]d = \frac{20}{n+1}[/tex]
And the arithmetic means are:
[tex]a, a + d, a + 2d, \ldots, a + nd[/tex]
Let's consider one more example.
Case 3: Inserting 4 Arithmetic Means
Insert (4) arithmetic means (n = 4):
[tex]20 = (4+1)d[/tex]
[tex]20 = 5d[/tex]
[tex]d = \frac{20}{5} = 4[/tex]
The sequence in this case is:
[tex]{-4, -4 + 4, -4 + 2 \cdot 4, -4 + 3 \cdot 4, -4 + 4 \cdot 4, 16}[/tex]
Simplifying:
[tex]-4, 0, 4, 8, 12, 16[/tex]
