(UEL) considere a progressão (-3,1,
...) o produto de seus 12 primeiros termos é:
a)
essa é a resposta quero a resolução
Soluções para a tarefa
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Cynthyaw, a fórmula que nos permite encontrar o produto dos termos de uma P.G finita é dada por:
Com efeito, devemos encontrar o último termo, isto é, o termo a_{12}. Segue,
![\\ \mathsf{a_n = a_1 \cdot q^{n - 1}} \\\\ \mathsf{a_{12} = - 3 \cdot \left ( - \frac{1}{3} \right )^{12 - 1}} \\\\ \mathsf{a_{12} = - 3 \cdot \left ( - \frac{1}{3} \right )^{11}} \\\\ \mathsf{a_{12} = - 3 \cdot \left [ \left ( - 3 \right )^{- 1} \right ]^{11}} \\\\ \mathsf{a_{12} = (- 3) \cdot ( - 3)^{- 11}} \\\\ \mathsf{a_{12} = (- 3)^{- 10}}](https://tex.z-dn.net/?f=%5C%5C+%5Cmathsf%7Ba_n+%3D+a_1+%5Ccdot+q%5E%7Bn+-+1%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7Ba_%7B12%7D+%3D+-+3+%5Ccdot+%5Cleft+%28+-+%5Cfrac%7B1%7D%7B3%7D+%5Cright+%29%5E%7B12+-+1%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7Ba_%7B12%7D+%3D+-+3+%5Ccdot+%5Cleft+%28+-+%5Cfrac%7B1%7D%7B3%7D+%5Cright+%29%5E%7B11%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7Ba_%7B12%7D+%3D+-+3+%5Ccdot+%5Cleft+%5B+%5Cleft+%28+-+3+%5Cright+%29%5E%7B-+1%7D+%5Cright+%5D%5E%7B11%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7Ba_%7B12%7D+%3D+%28-+3%29+%5Ccdot+%28+-+3%29%5E%7B-+11%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7Ba_%7B12%7D+%3D+%28-+3%29%5E%7B-+10%7D%7D "\\ \mathsf{a_n = a_1 \cdot q^{n - 1}} \\\\ \mathsf{a_{12} = - 3 \cdot \left ( - \frac{1}{3} \right )^{12 - 1}} \\\\ \mathsf{a_{12} = - 3 \cdot \left ( - \frac{1}{3} \right )^{11}} \\\\ \mathsf{a_{12} = - 3 \cdot \left [ \left ( - 3 \right )^{- 1} \right ]^{11}} \\\\ \mathsf{a_{12} = (- 3) \cdot ( - 3)^{- 11}} \\\\ \mathsf{a_{12} = (- 3)^{- 10}}")
Por fim, substituímos na fórmula...
![\\ \mathsf{P = \sqrt{(a_1 \cdot a_n)^n}} \\\\ \mathsf{P = \sqrt{\left [ (- 3) \cdot (- 3)^{- 10} \right ]^{12}}} \\\\ \mathsf{P = \left [ (- 3) \cdot (- 3)^{- 10} \right ]^{6}} \\\\ \mathsf{P = \left [ (- 3)^{- 9} \right ]^6} \\\\ \mathsf{P = (- 3)^{54}} \\\\ \boxed{\boxed{\mathsf{P = 3^{54}}}}](https://tex.z-dn.net/?f=%5C%5C+%5Cmathsf%7BP+%3D+%5Csqrt%7B%28a_1+%5Ccdot+a_n%29%5En%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7BP+%3D+%5Csqrt%7B%5Cleft+%5B+%28-+3%29+%5Ccdot+%28-+3%29%5E%7B-+10%7D+%5Cright+%5D%5E%7B12%7D%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7BP+%3D+%5Cleft+%5B+%28-+3%29+%5Ccdot+%28-+3%29%5E%7B-+10%7D+%5Cright+%5D%5E%7B6%7D%7D+%5C%5C%5C%5C+%5Cmathsf%7BP+%3D+%5Cleft+%5B+%28-+3%29%5E%7B-+9%7D+%5Cright+%5D%5E6%7D+%5C%5C%5C%5C+%5Cmathsf%7BP+%3D+%28-+3%29%5E%7B54%7D%7D+%5C%5C%5C%5C+%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7BP+%3D+3%5E%7B54%7D%7D%7D%7D "\\ \mathsf{P = \sqrt{(a_1 \cdot a_n)^n}} \\\\ \mathsf{P = \sqrt{\left [ (- 3) \cdot (- 3)^{- 10} \right ]^{12}}} \\\\ \mathsf{P = \left [ (- 3) \cdot (- 3)^{- 10} \right ]^{6}} \\\\ \mathsf{P = \left [ (- 3)^{- 9} \right ]^6} \\\\ \mathsf{P = (- 3)^{54}} \\\\ \boxed{\boxed{\mathsf{P = 3^{54}}}}")
Com efeito, devemos encontrar o último termo, isto é, o termo a_{12}. Segue,
Por fim, substituímos na fórmula...
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