From three number in GP other three number in GP are subtracted and the remainder are also found to be in GP prove that three sequences have the same common ratio.

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Solution

$Let the first three numbers in GP be : a_{1}, a_{1} r_{1}, a_{1} {r_{1}}^{2} and the other three numbers in GP be : a_{2}, a_{2} r_{2}, a_{2} {r_{2}}^{2} Numbers obtained after subtracting the numbers of second sequence from the first sequence are : = (a_{1} - a_{2}), (a_{1} r_{1} - a_{2} r_{2}), (a_{1} {r_{1}}^{2} - a_{2} {r_{2}}^{2}) Given that : the numbers obtained after subtraction are also in GP, thus : (a_{1} r_{1} - a_{2} r_{2})^{2} = (a_{1} - a_{2}) (a_{1} {r_{1}}^{2} - a_{2} {r_{2}}^{2}) {a_{1}}^{2} {r_{1}}^{2} + {a_{2}}^{2} {r_{2}}^{2} - 2 a_{1} a_{2} r_{1} r_{2} = {a_{1}}^{2} {r_{1}}^{2} - a_{1} a_{2} {r_{2}}^{2} - a_{1} a_{2} {r_{1}}^{2} + {a_{2}}^{2} {r_{2}}^{2} - 2 a_{1} a_{2} r_{1} r_{2} = - a_{1} a_{2} {r_{2}}^{2} - a_{1} a_{2} {r_{1}}^{2} 2 r_{1} r_{2} = {r_{1}}^{2} + {r_{2}}^{2} (r_{1} - r_{2})^{2} = 0 r_{1} = r_{2} Thus, the common ratio of the two sequences are same$

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