Question

Uma tropa de soldados é disposta em 20 fileiras, de forma que, em cada fileira, haja sempre x soldados a mais que na anterior. Nas 10 primeiras fileiras, há um total de 140 soldados; nas 10 últimas, 340 soldados. Calcule o valor de x, bem como o número de soldados na primeira e na última fila. Com resolução, por favo

Answer 1

f1=k

f2=k+x

...

fn=k+(n-1)x

 

$<var>\sum_{n=1}^{10}(k+(n-1)x)=140\\\\ 10*k + \sum_{n=1}^{10}(n-1)x=140\\\\ 10k+x*\sum_{n=1}^{10}(n-1)=140\\\\ 10k+x*\frac{0+9}{2}*10=140\\\\ 10k+x*45=140\\\\ 45x=140-10k\\\\ x=\frac{140-10k}{45}\\\\</var>$

 

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$<var>\sum_{n=11}^{20}(k+(n-1)x)=340\\\\ 10*k + \sum_{n=11}^{20}(n-1)x=340\\\\ 10k+x*\sum_{n=11}^{20}(n-1)=340\\\\ 10k+x*\frac{10+19}{2}*10=340\\\\ 10k+x*145=340\\\\ 145x=340-10k\\\\ x=\frac{340-10k}{145}\\\\ </var>$

 

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$<var>\begin{cases} x=\frac{140-10k}{45}\\x=\frac{340-10k}{145} \end{cases}\\\\\\ \frac{140-10k}{45}=\frac{340-10k}{145}\\\\ 145*(140-10k)=45*(340-10k)\\\\ 145*140-145*10k=45*340-45*10k\\\\ 20.300-1450k=15.300-450k\\\\ 20.300-15.300=-450k+1450k\\\\ 5.000=1000k\\\\ k=5</var>$

 

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$<var>x=\frac{140-10*5}{45}\\\\ x=\frac{140-50}{45}\\\\ x=\frac{90}{45}\\\\ x=2</var>$

 

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fn=k+(n-1)x

 

f1=5+(1-1)*2

f1=5

 

f20=5+(20-1)*2

f20=5+19*2

f20=5+38

f20=43

Answer 2

$<var>S_{10}=\frac{10 \cdot \left(a_1 + a_{10}\right)}{2}=140\\ 280=10 \cdot a_1 +10 \cdot a_{10}\right)\\ 10 \cdot a_1 =280-10 \cdot a_{10}\right)\\ a_1 =28- a_{10}\right)</var>$

$<var>a_{10}=28- a_1 \\a_{10} = a_1 + (10- 1) \cdot x \\28- a_1 = a_1 + (10- 1) \cdot x \\2a_1=-28+9x\\a_1=-14+4,5x</var>$