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




Question Number 135689 by joki last updated on 15/Mar/21
0,....,....,(5/4),(4/5),....  how to solution aritmatic
$$\mathrm{0},….,….,\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},…. \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{solution}\:\mathrm{aritmatic} \\ $$
Commented by liberty last updated on 15/Mar/21
AP ?
$${AP}\:? \\ $$
Commented by joki last updated on 15/Mar/21
common difference and sum
$$\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{sum}\: \\ $$$$ \\ $$
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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