Tens, Tens everywhere

Notice what happens when we multiply the number $10^{k+1}\sqrt{10}+10^{k}\sqrt{10}$ with $\dfrac{1}{10^{k+1}}\sqrt{10}+\dfrac{1}{10^{k}}\sqrt{10}$ : $(10^{k+1}\sqrt{10}+10^{k}\sqrt{10}) \left(\dfrac{1}{10^{k+1}}\sqrt{10}+\dfrac{1}{10^{k}}\sqrt{10}\right) = 10+100+1+10 = 121$ Thus, the geometric mean of those two numbers is $\sqrt{121}=11$ .

Since all of these numbers come in pairs, and the geometric mean of each pair is $11$ , the geometric mean of all the numbers is also $\boxed{11}$ .

the terms of the given expression can be simplified as /( 11\sqrt { 10 } \times { 10 }^{ 5 }\quad ;\quad 11\sqrt { 10 } \times { 10 }^{ 4 }\quad .............\quad 11\sqrt { 10 } \times { 10 }^{ -6 } /) . multiplying all these terms the above expression reduces to /( 11^{12} \times 10^{-6} \times 10^{6} /) which is /( 11^{12} /) . now taking twelfth root of /(11^{12} /) leaves us with /( \boxed{11} /) which is the answer