0,....,....,(5/4),(4/5),.... how to solution aritmatic


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

$0, . . . ., . . . ., \frac{5}{4}, \frac{4}{5}, . . . .$ $how to solution aritmatic$
Commented by liberty last updated on 15/Mar/21

$A P ?$
Commented by joki last updated on 15/Mar/21

$common difference and sum$
Commented by mr W last updated on 15/Mar/21

$i f i t i s a n A P, t h e n t h e c o m m o n$ $d i f f e r e n c e i s \frac{4}{5} - \frac{5}{4} = - \frac{9}{20}$ $t h a t m e a n s t h e A P i s d e c r e a s i n g!$ $a l l t e r m s b e f o r e t h e t e r m \frac{4}{5} m u s t$ $b e l a r g e r t h a n \frac{4}{5}, t h i s i s n o t t h e c a s e,$ $s i n c e t h e f i r s t t e r m i s 0, w h i c h i s$ $s m a l l e r t h a n \frac{4}{5} . t h e r e f o r e t h e$ $q u e s t i o n i s b a d, {i t}^{'} s n o t a n A P, b u t$ $w h a t i t i s, i s n o t c l e a r .$
Commented by MJS_new last updated on 15/Mar/21

$assuming 3 points given$ $P_{1} = (\begin{matrix} 1 \\ 0 \end{matrix}) P_{2} = (\begin{matrix} 4 \\ 5 / 4 \end{matrix}) P_{3} = (\begin{matrix} 5 \\ 4 / 5 \end{matrix})$ $y = {a x}^{2} + b x + c$ $(1) 0 = a + b + c$ $(2) \frac{5}{4} = 16 a + 4 b + c$ $(3) \frac{4}{5} = 25 a + 5 b + c$ $solving this system I get$ $a = - \frac{13}{60} \land b = \frac{3}{2} \land c = - \frac{77}{60}$ $y = - \frac{13}{60} x^{2} + \frac{3}{2} x - \frac{77}{60}$
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