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Given that

The number of terms in G.P. \[n = 4\]

Sum of the terms in G.P. = 820

Product of the terms in G.P. = 531441

Let \[a\] be the first term and \[r\] is the common ratio of the series of terms in G.P.

Let the first four terms be \[\dfrac{a}{{{r^3}}},\dfrac{a}{r},ar,a{r^3}\]

Now, consider the sum of the four terms

\[

\Rightarrow \dfrac{a}{{{r^3}}} + \dfrac{a}{r} + ar + a{r^3} = 820 \\

\Rightarrow a\left[ {\dfrac{1}{{{r^3}}} + \dfrac{1}{r} + r + {r^3}} \right] = 820.......................\left( 1 \right) \\

\]

And the product of the four terms is given by

\[ \Rightarrow \dfrac{a}{{{r^3}}} \times \dfrac{a}{r} \times ar \times a{r^3} = 531411\]

Cancelling the common terms, we have

\[

\Rightarrow a \times a \times a \times a = 531411 \\

\Rightarrow {a^4} = {\left( {27} \right)^4} \\

\therefore a = 27 \\

\]

Substituting \[a = 27\] in equation (1), we have

\[ \Rightarrow 27\left[ {\dfrac{1}{{{r^3}}} + \dfrac{1}{r} + r + {r^3}} \right] = 820\]

By trial and error method, let \[r = 3\]

\[

\Rightarrow 27\left[ {\dfrac{1}{{{3^3}}} + \dfrac{1}{3} + 3 + {3^3}} \right] = 820 \\

\Rightarrow 27\left[ {\dfrac{1}{{27}} + \dfrac{1}{3} + 3 + 27} \right] = 820 \\

\Rightarrow \dfrac{{27}}{{27}} + \dfrac{{27}}{3} + 27 \times 3 + 27 \times 27 = 820 \\

\Rightarrow 1 + 9 + 81 + 729 = 820 \\

\Rightarrow 820 = 820 \\

\]

Hence, \[r = 3\] satisfies the obtained equation.

Therefore, the four numbers are

\[

\Rightarrow \dfrac{a}{{{r^3}}} = \dfrac{{27}}{{{3^3}}} = \dfrac{{27}}{{27}} = 1 \\

\Rightarrow \dfrac{a}{r} = \dfrac{{27}}{3} = 9 \\

\Rightarrow ar = 27 \times 3 = 81 \\

\Rightarrow a{r^3} = 27 \times {3^3} = 27 \times 27 = 729 \\

\]