Three numbers, of which the third is equal to 36, form a geometric progression. If 36 is replaced with 20, then the three numbers form an arithmetic progression. Find these three numbers.

[tex]b-a=20-b\\\\\Rightarrow 2b=20+a\\\\\Rightarrow 2b=20+\dfrac{b^2}{36}\\\\\Rightarrow 72b=720+b^2\\\\\Rightarrow b^2-72b+720\\\\\Rightarrow b^2-60b-12b+720=0\\\\\Rightarrow b(b-60)-12(b-60)=0\\\\\Rightarrow (b-12)(b-60)\\\\\Rightarrow b=12,~~b=60.[/tex]

Therefore,

[tex]a=\dfrac{12^2}{36}=4,~~a=\dfrac{60^2}{36}=100.[/tex]

Thus, the three numbers are either 4, 12, 36 or 100, 60, 36.

aksnkj aksnkj · Answer 2 · 2021-11-15T07:16:12+01:00

The geometric series could be 4, 12, 36 or 100, 60, 36. And the arithmetic series will be 4, 12, 20 or 100, 60, 20.

Given information:

Three numbers, of which the third is equal to 36, form a geometric progression.

If 36 is replaced with 20, then the three numbers form an arithmetic progression.

Let the numbers be a, b, and 36 or 20 (for geometric or arithmetic series).

So, the third term 36 of geometric progression can be written as,

[tex]a_3=ar^2=36[/tex]

And the third term 20 of the arithmetic progression can be written as,

[tex]a_3'=a+2d[/tex]

Now, the geometric mean could be written as,

[tex]b^2=ac\\b^2=36a[/tex]

The arithmetic mean could be written as,

[tex]2b=a+c\\2b=a+20[/tex]

From the equation of geometric and arithmetic mean, it can be written,

[tex]b^2=36(2b-20)\\b^2-72b+720=0[/tex]

Solve the above quadratic equation for b as,

[tex]b^2-12b-60b+720=0\\b(b-12)-60(b-12)=0\\(b-12)(b-60)=0\\b=12, 60[/tex]

So, the value of middle term b can be 12 or 60.

Now, the first term a should be,

[tex]a=2b-20\\a=2\times 12-20=4\\\rm or\\a=2\times 60 -20=100[/tex]

Therefore, the geometric series could be 4, 12, 36 or 100, 60, 36. And the arithmetic series will be 4, 12, 20 or 100, 60, 20.

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https://brainly.com/question/13388802