Let number are a,ar,ar2,ar3
ar−1(r4−1)=1265…(1)
a1r1−1(r41−1)=1865
ar31(1−r1−r4)=1865…(2)
(2)(1)⇒a2r3=23
and a3⋅r3=1
ar=1
(ar)2⋅r=23
r=23,a=32
So, third term=ar2=32×49
α=23
2α=3
The sum of first four terms of a geometric progression (G.P.) is 1265 and the sum of their respective reciprocals is 1865. If the product of first three terms of the G.P. is 1, and the third term is α, then 2α is _________.
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