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Semenov [28]
3 years ago
6

1.) Three numbers form a geometric sequence whose common ratio is 0.5. If the first is reduced to 10 more than one quarter its v

alue, the second decreased by 10, and the third increased by 10 more than twice its value, the resulting three numbers form an arithmetic sequence. Determine the original three numbers.
Mathematics
1 answer:
emmainna [20.7K]3 years ago
8 0

Let <em>x</em> be the first number in the sequence, so the first three numbers are

{<em>x</em>, 0.5<em>x</em>, 0.5²<em>x</em>}

Then

{<em>x</em>/4 + 10, 0.5<em>x</em> - 10, 2(0.5²<em>x</em>) + 10}

is arithmetic, so there is some constant<em> c</em> such that

0.5<em>x</em> - 10 = <em>x</em>/4 + 10 + <em>c</em>   ==>   <em>x</em>/2 - 10 = <em>x</em>/4 + 10 + <em>c</em>

2(0.5²<em>x</em>) + 10 = 0.5<em>x</em> - 10 + <em>c</em>   ==>   <em>x</em>/2 + 10 = <em>x</em>/2 - 10 + <em>c</em>

Solve the second equation for <em>c</em> :

<em>x</em>/2 + 10 = <em>x</em>/2 - 10 + <em>c</em>

<em>c</em> = 20

Substitute this into the first equation and solve for <em>x</em> :

<em>x</em>/2 - 10 = <em>x</em>/4 + 10 + 20

<em>x</em>/4 = 40

<em>x</em> = 160

Then the terms are

{160, 80, 40}

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