Question

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Answer 1

Let the required numbers be \[a, \text { ar and a } r^2 .\]

Sum of the numbers = 21

\[\Rightarrow a + ar + a r^2 = 21\]

\[ \Rightarrow a(1 + r + r^2 ) = 21 . . . (i)\]

Sum of the squares of the numbers = 189

\[\Rightarrow a^2 + (ar )^2 + (a r^2 )^2 = 189\]

\[\Rightarrow a^2 + (ar )^2 + (a r^2 )^2 = 189 \]

\[ \Rightarrow a^2 \left( 1 + r^2 + r^4 \right) = 189 . . . (ii)\]

\[\text { Now }, a ( 1 + r + r^2 ) = 21 [\text { From } (i)]\]

\[\text { Squaring both the sides }\]

\[ \Rightarrow a^2 \left( 1 + r + r^2 \right)^2 = 441\]

\[ \Rightarrow a^2 \left( 1 + r^2 + r^4 \right) + 2 a^2 r\left( 1 + r + r^2 \right) = 441\]

\[ \Rightarrow 189 + 2ar\left\{ a\left( 1 + r + r^2 \right) \right\} = 441 [\text { Using } (ii)]\]

\[ \Rightarrow 189 + 2ar \times 21 = 441 [\text{ Using } (i)]\]

\[ \Rightarrow ar = 6\]

\[ \Rightarrow a = \frac{6}{r} . . . (iii)\]

\[\text { Putting } a = \frac{6}{r} \text { in }(i)\]

\[ \frac{6}{r}\left( 1 + r + r^2 \right) = 21\]

\[ \Rightarrow \frac{6}{r} + 6 + 6r = 21\]

\[ \Rightarrow 6 r^2 + 6r + 6 = 21r\]

\[ \Rightarrow 6 r^2 - 15r + 6 = 0\]

\[ \Rightarrow 3(2 r^2 - 5r + 2) = 0\]

\[ \Rightarrow 2 r^2 - 5r + 2 = 0\]

\[ \Rightarrow (2r - 1)(r - 2) = 0\]

\[ \Rightarrow r = \frac{1}{2}, 2\]

\[\text { Putting } r = \frac{1}{2} \text {in a } = \frac{6}{r}, \text { we get } a = 12 . \]

\[\text { So, the numbers are 12, 6 and 3 } . \]

\[\text { Putting } r = 2 in a = \frac{6}{r}, \text { we get a } = 3 . \]

\[\text { So, the numbers are 3, 6 and 12 } . \]

\[\text { Hence, the numbers that are in G . P are 3, 6 and 12 } . \]