quantos termos da p.g. (3; 6; 12;...) devemos somar a fim do que o total resulte 12285?
Soluções para a tarefa
Respondido por
3
Razão da PG a2/a1= 6/3 = 2
Sn=a1(qⁿ-1)/q-1
Substituindo na formula
12285=3(2ⁿ-1)/2-1
12285=3.(2ⁿ-1)
12284/3=2ⁿ-1
4095=2ⁿ-1
4095+1=2ⁿ
4096=2ⁿ
㏒4096=N㏒2
N=㏒4096/㏒2
N=3,61/0,30
N=12
Outra forma é fatorando o 4096 em primos veja
4096/2
2048/2
1024/2
512/2
256/2
128/2
64/2
32/2
16/2
8/2
4/2
2/2
1 = 2¹² = 4096
Igualando as bases
2¹²=2ⁿ
12=N
Logo temos que a PG tera 12 termos
A soma dos 12 termos será 12285
Espero ter ajudado!
Sn=a1(qⁿ-1)/q-1
Substituindo na formula
12285=3(2ⁿ-1)/2-1
12285=3.(2ⁿ-1)
12284/3=2ⁿ-1
4095=2ⁿ-1
4095+1=2ⁿ
4096=2ⁿ
㏒4096=N㏒2
N=㏒4096/㏒2
N=3,61/0,30
N=12
Outra forma é fatorando o 4096 em primos veja
4096/2
2048/2
1024/2
512/2
256/2
128/2
64/2
32/2
16/2
8/2
4/2
2/2
1 = 2¹² = 4096
Igualando as bases
2¹²=2ⁿ
12=N
Logo temos que a PG tera 12 termos
A soma dos 12 termos será 12285
Espero ter ajudado!
Respondido por
2
Boa tarde!
Solução!
![q= \dfrac{a2}{a1}= \dfrac{6}{3}=2\\\\\\
a1=3\\\\\
n=?\\\\\\\ S=12285](https://tex.z-dn.net/?f=q%3D+%5Cdfrac%7Ba2%7D%7Ba1%7D%3D+%5Cdfrac%7B6%7D%7B3%7D%3D2%5C%5C%5C%5C%5C%5C%0Aa1%3D3%5C%5C%5C%5C%5C%0An%3D%3F%5C%5C%5C%5C%5C%5C%5C+S%3D12285 "q= \dfrac{a2}{a1}= \dfrac{6}{3}=2\\\\\\
a1=3\\\\\
n=?\\\\\\\ S=12285")
Formula da soma de uma P.G!
![S= \dfrac{a1(q^{n}-1) }{q-1}\\\\\\\\
12285= \dfrac{3(2^{n}-1) }{2-1}\\\\\\\\
\dfrac{12285}{3}= \dfrac{(2^{n}-1) }{1}\\\\\\\\
4095=2^{n}-1\\\\\
4095+1=2^{n}\\\\\\
4096=2^{n}\\\\\\
Fazendo!\\\\\\
4096|2\\
2048|2\\
1024|2\\
~~~512|2\\
~~~256|2\\
~~~128|2\\
~~~~64|2\\
~~~~32|2\\
~~~~16|2\\
~~~~~8|2\\
~~~~~4|2\\
~~~~~2|2\\
~~~~~1\\\\\\\](https://tex.z-dn.net/?f=S%3D+%5Cdfrac%7Ba1%28q%5E%7Bn%7D-1%29+%7D%7Bq-1%7D%5C%5C%5C%5C%5C%5C%5C%5C%0A12285%3D+%5Cdfrac%7B3%282%5E%7Bn%7D-1%29+%7D%7B2-1%7D%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B12285%7D%7B3%7D%3D+%5Cdfrac%7B%282%5E%7Bn%7D-1%29+%7D%7B1%7D%5C%5C%5C%5C%5C%5C%5C%5C+%0A4095%3D2%5E%7Bn%7D-1%5C%5C%5C%5C%5C%0A4095%2B1%3D2%5E%7Bn%7D%5C%5C%5C%5C%5C%5C%0A4096%3D2%5E%7Bn%7D%5C%5C%5C%5C%5C%5C%0AFazendo%21%5C%5C%5C%5C%5C%5C%0A4096%7C2%5C%5C%0A2048%7C2%5C%5C%0A1024%7C2%5C%5C%0A%7E%7E%7E512%7C2%5C%5C%0A%7E%7E%7E256%7C2%5C%5C%0A%7E%7E%7E128%7C2%5C%5C%0A%7E%7E%7E%7E64%7C2%5C%5C%0A%7E%7E%7E%7E32%7C2%5C%5C%0A%7E%7E%7E%7E16%7C2%5C%5C%0A%7E%7E%7E%7E%7E8%7C2%5C%5C%0A%7E%7E%7E%7E%7E4%7C2%5C%5C%0A%7E%7E%7E%7E%7E2%7C2%5C%5C%0A%7E%7E%7E%7E%7E1%5C%5C%5C%5C%5C%5C%5C%0A%0A%0A%0A%0A+ "S= \dfrac{a1(q^{n}-1) }{q-1}\\\\\\\\
12285= \dfrac{3(2^{n}-1) }{2-1}\\\\\\\\
\dfrac{12285}{3}= \dfrac{(2^{n}-1) }{1}\\\\\\\\
4095=2^{n}-1\\\\\
4095+1=2^{n}\\\\\\
4096=2^{n}\\\\\\
Fazendo!\\\\\\
4096|2\\
2048|2\\
1024|2\\
~~~512|2\\
~~~256|2\\
~~~128|2\\
~~~~64|2\\
~~~~32|2\\
~~~~16|2\\
~~~~~8|2\\
~~~~~4|2\\
~~~~~2|2\\
~~~~~1\\\\\\\")
Logo!
![4096=2^{n}\\\\\
2^{12}=2^{n}\\\\\
\boxed{n=12}\\\\\\\
\boxed{Resp~~A~~P.G~~tem~~que~~ter~~12~~termos}](https://tex.z-dn.net/?f=4096%3D2%5E%7Bn%7D%5C%5C%5C%5C%5C%0A2%5E%7B12%7D%3D2%5E%7Bn%7D%5C%5C%5C%5C%5C%0A%5Cboxed%7Bn%3D12%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResp%7E%7EA%7E%7EP.G%7E%7Etem%7E%7Eque%7E%7Eter%7E%7E12%7E%7Etermos%7D++ "4096=2^{n}\\\\\
2^{12}=2^{n}\\\\\
\boxed{n=12}\\\\\\\
\boxed{Resp~~A~~P.G~~tem~~que~~ter~~12~~termos}")
Boa tarde!
Bons estudos!
Solução!
Formula da soma de uma P.G!
Logo!
Boa tarde!
Bons estudos!
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