Given AM of a1 and b1 is 165 with a>2. Since a,4,α,b are in AP:
4−a=α−4=b−α
This gives α=8−a and b=12−2a.
From AM condition: 2a1+b1=165 leads to aba+b=85
Substituting: 96−28a+2a220−3a=85
160−24a=480−140a+10a2
10a2−116a+320=0 or 5a2−58a+160=0
Using the quadratic formula gives a=4 or a=8.
For a=4: α=4,b=4.
The equation becomes 4x2−4x−8=0 or x2−x−2=0 with roots 2 and −1.
One root lies in (1,4) and the other in (−2,0).