3 numbers form an increasing GP . if the middle term is doubled, then the new numbers are in AP. find the common ratio of the GP
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♧♧HERE IS YOUR ANSWER♧♧
Let, the GP is
a, ar and ar²,
where a is the first term of the GP and r is the common ratio.
When, the middle term is double is doubled, we get an AP :
a, 2ar, ar²
So, the condition of AP suggests :
2nd term - 1st term = 3rd term - 2nd term
=> 2ar - a = ar² - 2ar
=> 2r - 1 = r² - 2r [eliminating a]
=> r² - 4r + 1 = 0
Using Sreedhar Acharya's method :
r = [- (-4) ± √{ (-4)² - 4.1.1 }]/(2.1)
= [4 ± √(16 - 4)]/2
= [4 ± √12]/2
= [4 ± 2√3]/2
= (2 ±√3)
So, the common ratio of the GP is (2 ±√3).
♧♧HOPE THIS HELPS♧♧
Let, the GP is
a, ar and ar²,
where a is the first term of the GP and r is the common ratio.
When, the middle term is double is doubled, we get an AP :
a, 2ar, ar²
So, the condition of AP suggests :
2nd term - 1st term = 3rd term - 2nd term
=> 2ar - a = ar² - 2ar
=> 2r - 1 = r² - 2r [eliminating a]
=> r² - 4r + 1 = 0
Using Sreedhar Acharya's method :
r = [- (-4) ± √{ (-4)² - 4.1.1 }]/(2.1)
= [4 ± √(16 - 4)]/2
= [4 ± √12]/2
= [4 ± 2√3]/2
= (2 ±√3)
So, the common ratio of the GP is (2 ±√3).
♧♧HOPE THIS HELPS♧♧
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