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Grade: 12th pass
        
Please solve this question as attached in the image. I am not getting the proper solution
one year ago

Answers : (1)

Aditya Gupta
1807 Points
							
so we can write fk= Sn – nak
or  fk/a=Sn/ak– n, where Sn = sum of a1, a2, …., an
or 1/αk = Sn/ak– n
now, 1/αk – 1/αk-1 = Sn*(1/ak – 1/ak-1)
but 1/ak – 1/ak-1= d (a constant coz a1, a2, …., an are in HP)
So, 1/αk – 1/αk-1 = Sn*d = constant
so α1, α2, α3, …. are in HP too
So, obviously 2^α1, 2^α2, 2^α3,.... , 2^αn are neither in GP (coz 2^αk /2^ αk-1 is not a constant), nor in AP, and also not in HP
so correct option is none of these.
one year ago
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