The sum of three number in GP is 42. If the first two number are increased by 2 and third is decreased by 4, the resulting form an AP, find the numbers of GP.

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$\{3, 9, 27\}$

${24, 12, 6}$

$\{64, 16, 4\}$

$\{12, 6, 3\}$

answer is A.

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Detailed Solution

Let three numbers be a, b and c

Given $a + b + c = 42$ …..(i)

Also, $a + 2, b + 2, c - 4 \in AP$

$\begin{array}{l} \Rightarrow a + 2 + c - 4 = 2 (b + 2) \\ \Rightarrow a + c - 2 = 2 b + 4 \end{array}$

$\Rightarrow a + c = 2 b + 6$ ... (ii)

From Eq. (i) and Eq. (ii), we ge

$\begin{array}{l} 2 b + 6 + b = 42 \\ \Rightarrow 3 b = 42 - 6 = 36 \\ b = 12 \\ \Rightarrow a r = 12 \end{array}$

From Eq. (i), we get

$\begin{array}{l} a + b + c = 42 \\ \Rightarrow a + a r + a r^{2} = 42 \\ \Rightarrow a + 12 + 12 r = 42 \\ \Rightarrow \frac{12}{r} + 12 + 12 r = 42 \\ \frac{1}{r} + 1 + r = \frac{42}{12} = \frac{7}{2} \\ \Rightarrow 2 (r^{2} + r + 1) = 7 r \\ \Rightarrow 2 r^{2} - 5 r + 2 = 0 \\ \Rightarrow 2 r^{2} - 4 r - r + 2 = 0 \\ \Rightarrow 2 r (r - 2) - (r - 2) = 0 \\ \Rightarrow (r - 2) (2 r - 1) = 0 \\ \Rightarrow r = 2, \frac{1}{2} \end{array}$