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Salut, am nevoie de ajutor la ex. 2.570B

Salut Am Nevoie De Ajutor La Ex 2570B class=

Răspuns :

Răspuns:

[tex]a_7 = 183[/tex]

Explicație pas cu pas:

[tex]a_5 = 61 = a_1\cdot q^4\\ \\ a_{11} = 1647 = a_1 \cdot q^{10}\\ \\ \frac{a_5}{a_11} = \frac{61}{1647} = \frac{a_1\cdot q^4}{a_1\cdot q^{10}}\\ \\ \frac{1}{27} = \frac{1}{q^6}\iff q^6 = 27\Bigg|\sqrt[3]{()}\\ \\ q^2 = \sqrt[3]{27} = 3\\ \\ q^2 = 3\iff q = \pm \sqrt{3}\\ \\ a_7 = a_1 \cdot q^6 = a_5\cdot q^2 \\ \\ \text{Pentru } q = \sqrt{3} \\ \\ a_7 = 61\cdot (\sqrt{3})^2 = 61\cdot 3 = 183\\ \\ \text{Pentru } q = -\sqrt{3}\\ \\ a_7 = 61\cdot (-\sqrt{3})^2 = 61 \cdot 3 = 183[/tex]

Formulă:

[tex]\boxed{a_n = a_k\cdot q^{n-k}}[/tex]

[tex]a_5 = 61 \quad(*)\\ a_{11} = 1647 \\ \\ \Rightarrow a_5\cdot q^6 = 1647\quad(**) \\ \\ \text{Impartim }(*)\text{ cu }(**)\Rightarrow q^6 = \dfrac{1647}{61} \Rightarrow q^6 = 27 \Rightarrow q = \pm\sqrt[6]{27} \Rightarrow \\ \\ \Rightarrow a_7 =a_5\cdot q^2\Rightarrow a_7= 61\cdot \Big(\pm\sqrt[6]{27}\Big)^2 = 61\cdot \sqrt[6]{27^2} =\\ \\=61\cdot \sqrt[6]{3^6} =61\cdot 3 = 183[/tex]

d) corect