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Matematyka 2.  Zakres podstawowy. Po gimnazjum

a) a1​=−7 a2​=a1​+3=−7+3=−4 a3​=a2​+3=−4+3=−1 a4​=a3​+3=−1+3=2 a5​=a4​+3=2+3=5 a6​=a5​+3=5+3=8 a7​=a6​+3=8+3=11     b) Jest to ciąg rosnący, ponieważ każdy następny wyraz powstaje przez zwiększenie poprzedniego wyrazu o 3.

Rozwiązanie 1

Ciąg jest arytmetyczny, jeśli różnica wyrazu o indeksie (n+1) i n jest stała (jest to r, czyli różnica ciągu arytmetycznego)   an+1​−an​=(−4(n+1)+17)−(−4n+17)=   =(−4n−4+17)+4n−17=   =−4n+13+4n−17=

a) {a1​=−3an​=an−1​+4      n&gt;1​

{(a13​=0    ∧   a13​=a1​+12r)   ⇒   a1​+12r=0(a29​=8   ∧   a29​=a1​+28r)   ⇒   a1​+28r=8​

Jeśli ten ciąg jest arytmetyczny, to wyraz środkowy jest średnią arytmetyczną dwóch pozostałych wyrazów.    2k=2(k2+k)+(k−3)​   ∣⋅2   4k=k2+k+k−3   4k=k2+2k−3   ∣−4k   k2−2k−3=0

a1​=3,   a2​,   a3​,   a4​,   a5​,   a6​=−9

Musimy sprawdzić, że wyraz środkowy jest średnią arytmetyczną dwóch pozostałych   7<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​−3)+3−7<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​2​]   ∣⋅2   27<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​ =? 7<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​−3+3−7<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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