Question

The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Answer 1

Let the numbers be a, ar and ar².

\[\text { Sum  }= 14 \]

\[ \Rightarrow a + ar + a r^2 = 14 \]

\[ \Rightarrow a(1 + r + r^2 ) = 14 . . . \left( i \right)\]

According to the question,a + 1, ar + 1 and ar² − 1 are  in A.P.

\[\therefore 2\left( ar + 1 \right) = a + 1 + a r^2 - 1\]

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

\[ \Rightarrow 2ar + 2 = 14 - ar [\text { From }\left( i \right)]\]

\[ \Rightarrow 3ar = 12 \]

\[ \Rightarrow a = \frac{4}{r} . . . \left( ii \right)\]

\[\text { Putting } a = \frac{4}{r} \text { in }\left( i \right)\]

\[ \Rightarrow \frac{4}{r}(1 + r + r^2 ) = 14\]

\[ \Rightarrow 4 r^2 - 10r + 4 = 0 \]

\[ \Rightarrow 4 r^2 - 8r - 2r + 4 = 0 \]

\[ \Rightarrow \left( 4r - 2 \right)\left( r - 2 \right) = 0\]

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

\[\text { Putting } r = \frac{1}{2}\text { in } \left( ii \right), \text { we get a } = 8 . \]

\[\text { So, the G . P . is 8, 4 and } 2 . \]

Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.