Let a,b,c>1,a^3,b^3 and ^3 be in A.P. and log_ab, log_ a and log_bc be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c 3…

Given,

${a}^{3},{b}^{3},{c}^{3}$a3,b3,c3 be in $A.P.$A.P.

So, ${a}^{3}+{c}^{3}=2{b}^{3}........(1)$a3+c3=2b3........(1)

Also given {\mathrm{log}}_{a}b,{\mathrm{log}}_{c}a&{\mathrm{log}}_{b}c are in $G.P.$G.P.

So, $\frac{\mathrm{log}b}{\mathrm{log}a}\cdot \frac{\mathrm{log}c}{\mathrm{log}b}={(\frac{\mathrm{log}a}{\mathrm{log}c})}^{2}$logalogb​⋅logblogc​=(logcloga​)2

$\Rightarrow (\mathrm{log}a{)}^{3}=(\mathrm{log}c{)}^{3}\Rightarrow a=c$⇒(loga)3=(logc)3⇒a=c ..........(2)

Now from equation (1)&(2) we get, $a=b=c$a=b=c

Now, ${T}_{1}=\frac{a+4b+c}{3}=2a$T1​=3a+4b+c​=2a and $d=\frac{a-8b+c}{10}=\frac{-6a}{10}=\frac{-3}{5}a$d=10a−8b+c​=10−6a​=5−3​a

$$\therefore {S}_{20}=\frac{20}{2}[4a+19(-\frac{3}{5}a)]$$∴S20​=220​[4a+19(−53​a)]

$$\Rightarrow {S}_{20}=10[\frac{20a-57a}{5}]$$⇒S20​=10[520a−57a​]

$$\Rightarrow {S}_{20}=-74a$$⇒S20​=−74a

$$\Rightarrow -444=-74a\Rightarrow a=6$$⇒−444=−74a⇒a=6

Hence, $abc={6}^{3}=216$abc=63=216