Question

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

Answer 1

Let the first term of a G.P be a and its common ratio be r.

\[\therefore a_1 + a_2 + a_3 = 56\]

\[ \Rightarrow a + ar + a r^2 = 56 \]

\[ \Rightarrow a \left( 1 + r + r^2 \right) = 56 \]

\[ \Rightarrow a = \frac{56}{1 + r + r^2} . . . . . . . \left( i \right) \]

\[\text { Now, according to the question }: \]

\[a - 1, ar - 7 \text { and }{ar}^2 - 21 \text { are in A . P } . \]

\[ \therefore 2\left( ar - 7 \right) = a - 1 + {ar}^2 - 21\]

\[ \Rightarrow 2ar - 14 = {ar}^2 + a - 22\]

\[ \Rightarrow {ar}^2 - 2ar + a - 8 = 0\]

\[ \Rightarrow a \left( 1 - r \right)^2 = 8\]

\[ \Rightarrow a = \frac{8}{\left( 1 - r \right)^2} . . . . . . . \left( ii \right)\]

\[\text { Equating (i) and  (ii) }: \]

\[ \Rightarrow \frac{8}{\left( 1 - r \right)^2} = \frac{56}{1 + r + r^2}\]

\[ \Rightarrow 8\left( 1 + r + r^2 \right) = 56\left( 1 + r^2 - 2r \right) \Rightarrow 1 + r + r^2 = 7 \left( 1 + r^2 - 2r \right)\]

\[ \Rightarrow 1 + r + r^2 = 7 + 7 r^2 - 14r\]

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

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

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

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

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

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

 \[ \text{ When r } = 2, a = 8 . [\text { Using } (ii)]\]

\[\text { And, the required numbers are 8, 16 and 32 } . \]

\[\text {When r } = \frac{1}{2}, a = 32 . [\text { Using } (ii)]\]

\[\text { And, the required numbers are 32, 16 and 8 }. \]