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Sagot :
[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
[tex] \textbf{Let's see if the sequence is Arithmetic or Geometric :} [/tex]
[tex] \textsf{If the ratio between successive terms is } [/tex] [tex] \textsf{equal then, the terms are in GP} [/tex]
- [tex] \sf{ \dfrac{12}{-4} = -3} [/tex]
- [tex] \sf{ \dfrac{-36}{12} = -3} [/tex]
[tex] \textsf{Since the common ratio is same, } [/tex] [tex] \textsf{we can infer that it's a geometric progression} [/tex] [tex] \textsf{with common ratio of -3} [/tex]
Answer:
okay
Step-by-step explanation:
[tex]for arithmetic \: progression \: there \: must \: be \: a \: common \: difference \\ \\ to \: get \: common \: difference \: {d} \: \\ get \: the \: the \: term \: infront \: minus \: the \: term \: before \\ say \: d = 12 - ( - 4) = 16 \\ d = - 36 - 12 = - 48 \\ d = 108 - ( - 36) = 144 \\ since \: d \: isnt \: uniform \: this \: is \: not \: arithmetic \: progression \\ \\ for \: geometric \: progression \: there \: must \: be \: a \: common \: ratio \: {r} \\ r \: is \: got \: by \: dividing \: the \: term \: infront \: by \: the \: term \: behind \\ \\ say \: r = 12 \div ( - 4) = - 3 \\ r = ( - 36) \div 12 = - 3 \\ r = 108 \div ( - 36) = - 3 \\ since \: r \: is \: the \: same \: this \: is \: a \: geometric \: serie[/tex]
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