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If-12-x-y-and-4-provides-a-sequence-such-that-the-first-3-of-the-numbers-are-in-arithmetic-progression-Calculate-the-a-Possible-values-of-x-and-y-b-The-sum-of-the-A-P-c-The-product-of-the-la

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Question Number 10318 by Tawakalitu ayo mi last updated on 03/Feb/17
If 12, x, y and 4 provides a sequence such that the first 3 of the numbers are in arithmetic progression. Calculate the (a) Possible values of x and y (b) The sum of the A.P (c) The product of the last 3 numbers of the G.P
$$\mathrm{If}\:\mathrm{12},\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{4}\:\mathrm{provides}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{3}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}.\:\mathrm{Calculate}\:\mathrm{the}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{A}.\mathrm{P} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{3}\:\mathrm{numbers}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{G}.\mathrm{P} \\ $$
Answered by mrW1 last updated on 05/Feb/17
(a) the first 3 numbers are in A.P. ⇒ y−x=x−12 (i) the last 3 numbers are in G.P.: ⇒(4/y)=(y/x) (ii) from (i) we get: y=2x−12=2(x−6) from (ii) we get: y^2 =4x 4(x^2 −12x+36)=4x x^2 −13x+36=0 x=((13±(√(13^2 −4×36)))/2)=((13±5)/2)= { (9),(4) :} y=2(x−6)= { (6),((−4)) :} the sequence is 12, 9, 6, 4 or 12, 4, −4, 4 (b) 12+9+6=27 or 12+4−4=12 (c) 9×6×4=216 or 4×(−4)×4=−64
$$\left({a}\right) \\ $$$${the}\:{first}\:\mathrm{3}\:{numbers}\:{are}\:{in}\:{A}.{P}. \\ $$$$\Rightarrow\:{y}−{x}={x}−\mathrm{12}\:\:\:\:\:\left({i}\right) \\ $$$${the}\:{last}\:\mathrm{3}\:{numbers}\:{are}\:{in}\:{G}.{P}.: \\ $$$$\Rightarrow\frac{\mathrm{4}}{{y}}=\frac{{y}}{{x}}\:\:\:\:\:\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{we}\:{get}: \\ $$$${y}=\mathrm{2}{x}−\mathrm{12}=\mathrm{2}\left({x}−\mathrm{6}\right) \\ $$$$ \\ $$$${from}\:\left({ii}\right)\:{we}\:{get}: \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x} \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{36}\right)=\mathrm{4}{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{36}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{13}\pm\sqrt{\mathrm{13}^{\mathrm{2}} −\mathrm{4}×\mathrm{36}}}{\mathrm{2}}=\frac{\mathrm{13}\pm\mathrm{5}}{\mathrm{2}}=\begin{cases}{\mathrm{9}}\\{\mathrm{4}}\end{cases} \\ $$$${y}=\mathrm{2}\left({x}−\mathrm{6}\right)=\begin{cases}{\mathrm{6}}\\{−\mathrm{4}}\end{cases} \\ $$$$ \\ $$$${the}\:{sequence}\:{is} \\ $$$$\mathrm{12},\:\mathrm{9},\:\mathrm{6},\:\mathrm{4}\:{or} \\ $$$$\mathrm{12},\:\mathrm{4},\:−\mathrm{4},\:\mathrm{4} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\mathrm{12}+\mathrm{9}+\mathrm{6}=\mathrm{27}\:{or} \\ $$$$\mathrm{12}+\mathrm{4}−\mathrm{4}=\mathrm{12} \\ $$$$ \\ $$$$\left({c}\right) \\ $$$$\mathrm{9}×\mathrm{6}×\mathrm{4}=\mathrm{216}\:{or} \\ $$$$\mathrm{4}×\left(−\mathrm{4}\right)×\mathrm{4}=−\mathrm{64} \\ $$
Commented by Tawakalitu ayo mi last updated on 04/Feb/17
I really appreciate your effort sir. God bless you sir.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless} \\ $$$$\mathrm{you}\:\mathrm{sir}. \\ $$

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