Consider x - y - z are three consecutive terms of a G-P- such that x - y - z -120 and y -80- Find xyz- -A- 512000B- 1024000C- 256000D- 800000

The correct option is A 512000 Let r be the common ratio of the G.P. 80/r + 80 + 80r = 120 Solving we get r = 2 or ( 1/2 ) So x and z will be 160 and 40 ...

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Question

Consider x, y, z are three consecutive terms of a G.P. such that x + y + z = -120 and y = 80. Find xyz. ?

256000

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800000

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512000

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1024000

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Solution

The correct option is C
512000

Let r be the common ratio of the G.P.

$\frac{80}{r}$ + 80 + 80r = -120

Solving we get r = -2 or - $(\frac{1}{2})$

So x and z will be -160 and -40 or -40 and -160 respectively.

xyz = 512000

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Consider x, y, z are three consecutive terms of a G.P. such that x + y + z = -120 and y = 80. Find xyz. ?

The correct option is C 512000 Let r be the common ratio of the G.P. 80r + 80 + 80r = -120 Solving we get r = -2 or - (12) So x and z will be -160 and -40 or -40 and -160 respectively. xyz = 512000

The correct option is C
512000

Let r be the common ratio of the G.P.

$\frac{80}{r}$ + 80 + 80r = -120

Solving we get r = -2 or - $(\frac{1}{2})$

So x and z will be -160 and -40 or -40 and -160 respectively.

xyz = 512000