An algebra problem by Anuj Shikarkhane

The sum of the first 5 terms of a geometric sequence can be written as $x+ax+{ a }^{ 2 }x+{ a }^{ 3 }x+{ a }^{ 4 }x$ .

Because the product of the first 3 terms is ${ 9 }^{ 6 }$ , that means that $x\cdot ax\cdot { a }^{ 2 }x={ 9 }^{ 6 }$ , or ${ { a }^{ 3 }x }^{ 3 }={ 3 }^{ 12 }$ .

Similarly, because the product of the last 3 terms is ${ 3 }^{ 6 }$ , that means that ${ a }^{ 2 }x\cdot { a }^{ 3 }x\cdot { a }^{ 4 }x={ 3 }^{ 6 }$ , or ${ a }^{ 9 }{ x }^{ 3 }={ 3 }^{ 6 }$ .

Dividing the two equations, we get $\frac { { a }^{ 9 }{ x }^{ 3 } }{ { a }^{ 3 }{ x }^{ 3 } } =\frac { { 3 }^{ 6 } }{ { 3 }^{ 12 } }$ , or $a=\frac { 1 }{ { 3 } }$ .

Replacing $a$ with $\frac { 1 }{ 3 }$ in one of the equations, we get that $x=3^{ 5 }$ .

Plugging these values into the original sequence, we get that $3^{ 5 }+\frac { 3^{ 5 } }{ 3^{ 1 } } +\frac { 3^{ 5 } }{ 3^{ 2 } } +\frac { 3^{ 5 } }{ 3^{ 3 } } +\frac { 3^{ 5 } }{ 3^{ 4 } } =\boxed { 363 }$ .