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If x-6,x-2 and x+10 are three consecutive terms of a geometric progression,find the value of x​

User Muposat
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1 Answer

15 votes
15 votes

Explanation:

let the common ratio be r

(x-6)r=(x-2)

(x-2)r=(x+10)

r = (x-2)/(x-6)

and r = (x+10)/(x-2)

so, (x-2)/(x-6)=(x+10)/(x-2)

sloving it, you'll get x=8

so the value of x is 8

User Johnykes
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