0-5-4-4-5-how-to-solution-aritmatic-

Question Number 135689 by joki last updated on 15/Mar/21

$$\mathrm{0},….,….,\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},…. \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{solution}\:\mathrm{aritmatic} \\ $$

Commented by liberty last updated on 15/Mar/21

$${AP}\:? \\ $$

Commented by joki last updated on 15/Mar/21

$$\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{sum}\: \\ $$$$ \\ $$

Commented by mr W last updated on 15/Mar/21

if it is an AP, then the common difference is (4/5)−(5/4)=−(9/(20)) that means the AP is decreasing! all terms before the term (4/5) must be larger than (4/5), this is not the case, since the first term is 0, which is smaller than (4/5). therefore the question is bad, it′s not an AP, but what it is, is not clear.

$${if}\:{it}\:{is}\:{an}\:{AP},\:{then}\:{the}\:{common} \\ $$$${difference}\:{is}\:\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{4}}=−\frac{\mathrm{9}}{\mathrm{20}} \\ $$$${that}\:{means}\:{the}\:{AP}\:{is}\:{decreasing}! \\ $$$${all}\:{terms}\:{before}\:{the}\:{term}\:\frac{\mathrm{4}}{\mathrm{5}}\:{must} \\ $$$${be}\:{larger}\:{than}\:\frac{\mathrm{4}}{\mathrm{5}},\:{this}\:{is}\:{not}\:{the}\:{case}, \\ $$$${since}\:{the}\:{first}\:{term}\:{is}\:\mathrm{0},\:{which}\:{is} \\ $$$${smaller}\:{than}\:\frac{\mathrm{4}}{\mathrm{5}}.\:{therefore}\:{the} \\ $$$${question}\:{is}\:{bad},\:{it}'{s}\:{not}\:{an}\:{AP},\:{but} \\ $$$${what}\:{it}\:{is},\:{is}\:{not}\:{clear}. \\ $$

Commented by MJS_new last updated on 15/Mar/21

assuming 3 points given P_1 = ((1),(0) ) P_2 = ((4),((5/4)) ) P_3 = ((5),((4/5)) ) y=ax^2 +bx+c (1) 0=a+b+c (2) (5/4)=16a+4b+c (3) (4/5)=25a+5b+c solving this system I get a=−((13)/(60))∧b=(3/2)∧c=−((77)/(60)) y=−((13)/(60))x^2 +(3/2)x−((77)/(60))

$$\mathrm{assuming}\:\mathrm{3}\:\mathrm{points}\:\mathrm{given} \\ $$$${P}_{\mathrm{1}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:\:\:{P}_{\mathrm{2}} =\begin{pmatrix}{\mathrm{4}}\\{\mathrm{5}/\mathrm{4}}\end{pmatrix}\:\:\:{P}_{\mathrm{3}} =\begin{pmatrix}{\mathrm{5}}\\{\mathrm{4}/\mathrm{5}}\end{pmatrix} \\ $$$${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{0}={a}+{b}+{c} \\ $$$$\left(\mathrm{2}\right)\:\:\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{16}{a}+\mathrm{4}{b}+{c} \\ $$$$\left(\mathrm{3}\right)\:\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{25}{a}+\mathrm{5}{b}+{c} \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{system}\:\mathrm{I}\:\mathrm{get} \\ $$$${a}=−\frac{\mathrm{13}}{\mathrm{60}}\wedge{b}=\frac{\mathrm{3}}{\mathrm{2}}\wedge{c}=−\frac{\mathrm{77}}{\mathrm{60}} \\ $$$${y}=−\frac{\mathrm{13}}{\mathrm{60}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{x}−\frac{\mathrm{77}}{\mathrm{60}} \\ $$

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