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Sum-of-three-numbers-in-GP-be-14-If-one-is-added-to-first-and-second-and-1-is-subtracted-from-the-third-the-new-numbers-are-in-AP-The-smallest-of-them-is-

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Question Number 12827 by 786786AM last updated on 03/May/17
Sum of three numbers in GP be 14. If one is added to first and second and 1 is subtracted from the third, the new numbers are in AP. The smallest of them is
$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{be}\:\mathrm{14}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{is} \\ $$$$\mathrm{added}\:\mathrm{to}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{and}\:\mathrm{1}\:\mathrm{is}\:\mathrm{subtracted} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{third},\:\mathrm{the}\:\mathrm{new}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}. \\ $$$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{of}\:\mathrm{them}\:\mathrm{is} \\ $$
Answered by mrW1 last updated on 04/May/17
a_1 =a a_2 =aq a_3 =aq^2 a_1 +a_2 +a_3 =14 a(1+q+q^2 )=14 (a_2 +1)−(a_1 +1)=(a_3 −1)−(a_2 +1) 2(a_2 +1)=(a_1 +1)+(a_3 −1) 2(aq+1)=a+1+aq^2 −1 aq^2 −2aq+a−2=0 aq^2 +aq+a−3aq−2=0 14−3aq−2=0 aq=4 a+aq+aq^2 =a+4+((16)/a)=14 a+((16)/a)=10 a^2 −10a+16=0 a=((10±(√(100−4×16)))/2)=((10±6)/2)=8 or 2 q=(1/2) or 2 the numbers are 2, 4, 8 or 8, 4, 2 ⇒the smallest of them is 2.
$${a}_{\mathrm{1}} ={a} \\ $$$${a}_{\mathrm{2}} ={aq} \\ $$$${a}_{\mathrm{3}} ={aq}^{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} =\mathrm{14} \\ $$$${a}\left(\mathrm{1}+{q}+{q}^{\mathrm{2}} \right)=\mathrm{14} \\ $$$$ \\ $$$$\left({a}_{\mathrm{2}} +\mathrm{1}\right)−\left({a}_{\mathrm{1}} +\mathrm{1}\right)=\left({a}_{\mathrm{3}} −\mathrm{1}\right)−\left({a}_{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{2}\left({a}_{\mathrm{2}} +\mathrm{1}\right)=\left({a}_{\mathrm{1}} +\mathrm{1}\right)+\left({a}_{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\mathrm{2}\left({aq}+\mathrm{1}\right)={a}+\mathrm{1}+{aq}^{\mathrm{2}} −\mathrm{1} \\ $$$${aq}^{\mathrm{2}} −\mathrm{2}{aq}+{a}−\mathrm{2}=\mathrm{0} \\ $$$${aq}^{\mathrm{2}} +{aq}+{a}−\mathrm{3}{aq}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{14}−\mathrm{3}{aq}−\mathrm{2}=\mathrm{0} \\ $$$${aq}=\mathrm{4} \\ $$$${a}+{aq}+{aq}^{\mathrm{2}} ={a}+\mathrm{4}+\frac{\mathrm{16}}{{a}}=\mathrm{14} \\ $$$${a}+\frac{\mathrm{16}}{{a}}=\mathrm{10} \\ $$$${a}^{\mathrm{2}} −\mathrm{10}{a}+\mathrm{16}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4}×\mathrm{16}}}{\mathrm{2}}=\frac{\mathrm{10}\pm\mathrm{6}}{\mathrm{2}}=\mathrm{8}\:{or}\:\mathrm{2} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{2} \\ $$$${the}\:{numbers}\:{are}\:\mathrm{2},\:\mathrm{4},\:\mathrm{8}\:{or}\:\mathrm{8},\:\mathrm{4},\:\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow{the}\:{smallest}\:{of}\:{them}\:{is}\:\mathrm{2}. \\ $$
Answered by sandy_suhendra last updated on 04/May/17
the other way, we can start from AP AP : (a−b), a , (a+b) GP : (a−b−1) , (a−1) , (a+b+1) (a−b−1)+(a−1)+(a+b+1)=14 3a−1=14 3a=15 ⇒ a=5 GP : (4−b) , 4 , (6+b) (4−b)(6+b)=4^2 24−2b−b^2 =16 b^2 +2b−8=0 (b+4)(b−2)=0 b=−4 ⇒ 8,4,2 b=2 ⇒2,4,8
$$\mathrm{the}\:\mathrm{other}\:\mathrm{way},\:\mathrm{we}\:\mathrm{can}\:\mathrm{start}\:\mathrm{from}\:\mathrm{AP}\:\:\:\:\: \\ $$$$\mathrm{AP}\::\:\left(\mathrm{a}−\mathrm{b}\right),\:\mathrm{a}\:,\:\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{GP}\::\:\left(\mathrm{a}−\mathrm{b}−\mathrm{1}\right)\:,\:\left(\mathrm{a}−\mathrm{1}\right)\:,\:\left(\mathrm{a}+\mathrm{b}+\mathrm{1}\right) \\ $$$$\left(\mathrm{a}−\mathrm{b}−\mathrm{1}\right)+\left(\mathrm{a}−\mathrm{1}\right)+\left(\mathrm{a}+\mathrm{b}+\mathrm{1}\right)=\mathrm{14} \\ $$$$\mathrm{3a}−\mathrm{1}=\mathrm{14} \\ $$$$\mathrm{3a}=\mathrm{15}\:\Rightarrow\:\mathrm{a}=\mathrm{5} \\ $$$$\mathrm{GP}\::\:\left(\mathrm{4}−\mathrm{b}\right)\:,\:\mathrm{4}\:,\:\left(\mathrm{6}+\mathrm{b}\right) \\ $$$$\left(\mathrm{4}−\mathrm{b}\right)\left(\mathrm{6}+\mathrm{b}\right)=\mathrm{4}^{\mathrm{2}} \\ $$$$\mathrm{24}−\mathrm{2b}−\mathrm{b}^{\mathrm{2}} =\mathrm{16} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{2b}−\mathrm{8}=\mathrm{0} \\ $$$$\left(\mathrm{b}+\mathrm{4}\right)\left(\mathrm{b}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{b}=−\mathrm{4}\:\Rightarrow\:\mathrm{8},\mathrm{4},\mathrm{2} \\ $$$$\mathrm{b}=\mathrm{2}\:\Rightarrow\mathrm{2},\mathrm{4},\mathrm{8} \\ $$

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