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The sum of four numbers in G.P. is 820 and their product is \[5,31,441\]. Find the numbers.

Answer Verified Verified
Hint: In this question, first of all consider the four numbers as \[\dfrac{a}{{{r^3}}},\dfrac{a}{r},ar,a{r^3}\] which are in G.P. Then equate the sum and product of these terms to the given sum and product of the four numbers which will give us the unknown values.

Complete step-by-step answer:
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
\[
   \Rightarrow \dfrac{a}{{{r^3}}} = \dfrac{{27}}{{{3^3}}} = \dfrac{{27}}{{27}} = 1 \\
   \Rightarrow \dfrac{a}{r} = \dfrac{{27}}{3} = 9 \\
   \Rightarrow ar = 27 \times 3 = 81 \\
   \Rightarrow a{r^3} = 27 \times {3^3} = 27 \times 27 = 729 \\
\]
Thus, the four numbers which are in G.P. are \[1,9,81,729\]

Note: In these kinds of questions, consider the terms as \[\dfrac{a}{r},a,ar\] whenever three terms are in G.P. and consider the terms as \[\dfrac{a}{{{r^3}}},\dfrac{a}{r},ar,a{r^3}\] whenever four terms are in G.P.