Given,
a3,b3,c3 be in A.P.
So, a3+c3=2b3........(1)
Also given {\mathrm{log}}_{a}b,{\mathrm{log}}_{c}a&{\mathrm{log}}_{b}c are in G.P.
So, logalogb⋅logblogc=(logcloga)2
⇒(loga)3=(logc)3⇒a=c ..........(2)
Now from equation (1)&(2) we get, a=b=c
Now, T1=3a+4b+c=2a and d=10a−8b+c=10−6a=5−3a
∴S20=220[4a+19(−53a)]
⇒S20=10[520a−57a]
⇒S20=−74a
⇒−444=−74a⇒a=6
Hence, abc=63=216