Mathematics - Sequence & Series Question with Solution | TestHub

$a,b,c$ are in A.P.

$b^2=ac$

$$\begin{array}{c}{\log }_{6}a+{\log }_{6}b+{\log }_{6}c=6abc=6^6{b}^3=6^{6}b=6^2=36\end{array}$$

$\mathrm{ac}=36\times 36=2^4\times 3^4$ $\mathrm{b}-\mathrm{a}={\mathrm{N}}^2$ $36-\mathrm{a}={\mathrm{N}}^2$ a is factor of $2^4{3}^4$ $\mathrm{a}=27$ is possible value $36-27=9=(3{)}^2$

$$\begin{array}{rlr}\Rightarrow & a=27,b=36,c=48& A+b+c=111\text{ Ans. }\end{array}$$