If first three terms of the sequence `1//16 ,a , b , 1/6` are in geometric series and last three terms are in harmonic series, then find the values of `a a n d bdot`

Explanation

1/16, a,b are in G.P. Hence,
`a^(2)=b/16or16a^(2)=b` (1)
`rArra,b,1/6` are in H.P. Hence,
`b=(2a1/6)/(a+1/6)=(2a)/(6a+1)` (2)
From (1) and (2),
`16a^(2)=(2a)/(6a+1)`
or `2a(8a-1/(6a+1))=0`
or 8a(6a+1)-1=0
or `48a^(2)+8a-1=0` (`becausene0)`
or(4a+1)(12a-1)=0
`thereforea=-1/4,1/12`
when a=-1/4, then from (1),
`b=16(-1/4)^(2)=1` ltbr .When a=1/12, then from (1),
`b=16(1/12)^(2)=1/9`.
Therefore,a=-1/4,b=1ora=1/12,b=1/9.