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Sagot :
The geometric and arithmetic progression are a number series such that
each term is given by adding or multiplying the previous term by a constant.
Responses:
The possible three numbers are;
- 4, 8, 16
- [tex]\displaystyle \frac{4}{25} , \ -\frac{16}{25}, \ \frac{64}{25}[/tex]
Methods by which the above responses are obtained:
Let the three numbers be a, a·r, and a·r², we have;
Given;
G.P. = a, a·r, a·r²
A.P. = a, a·r + 2, a·r²
G.P. = a, a·r + 2, a·r² + 9
Solution:
(a·r + 2) - a = a·r² - (a·r + 2) by definition of an AP
a·r + 2 - a = a·r² - (a·r + 2)
(r² - 2·r + 1)·a - 4 = 0
G.P = a, a·r + 2, a·r² + 9
[tex]\displaystyle \frac{a \cdot r + 2}{a} = \frac{a \cdot r^2 + 9}{a \cdot r + 2}[/tex]
(a·r + 2)² = a·(a·r² + 9)
r²·a² + 4·r·a + 4 = r²·a² + 9·a
4·r·a + 4 = 9·a
4 = 9·a - 4·r·a = a·(9 - 4·r)
[tex]\displaystyle a = \mathbf{ \frac{4}{9 - 4 \cdot r}}[/tex]
[tex]\displaystyle \left(r^2 - 2 \cdot r + 1 \right) \times \frac{4}{9 - 4 \cdot r} - 4 = 0[/tex]
Therefore;
[tex]\displaystyle \left(r^2 - 2 \cdot r + 1 \right) \times \frac{4}{9 - 4 \cdot r} \times (9 - 4 \cdot r) - 4 \times (9 - 4 \cdot r)= 0[/tex]
4·r² + 8·r - 32 = 0
Which gives;
4·r² + 8·r - 32 = 4·(r - 2)·(r + 4) = 0
r = 2 or r = -4
[tex]\displaystyle a = \frac{4}{9 - 4 \times 2} = \mathbf{ 4}[/tex] or [tex]\displaystyle a = \frac{4}{9 - 4 \times (-4)} = \frac{4}{25}[/tex]
The first three numbers are;
4, 4 × 2, 4 × 2², which gives;
- 4, 8, 16
[tex]\displaystyle When \ a = \frac{4}{25}, and \ r = -4, we \ have;[/tex]
[tex]\displaystyle \frac{4}{25}, \ \frac{4}{25} \times (-4), \ \frac{4}{25} \times (-4)^2[/tex]
Which gives;
[tex]\displaystyle \frac{4}{25} , \ -\frac{16}{25} , \ \frac{64}{25}[/tex]
[tex]\displaystyle The \ second \ three \ numbers \ are; \underline{\frac{4}{25}, \ -\frac{16}{25} , \ \frac{64}{25}}[/tex]
Learn more about geometric and arithmetic progression here:
https://brainly.com/question/1800090
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