Resposta :
f1=k
f2=k+x
...
fn=k+(n-1)x
[tex]\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}\\\\[/tex]
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[tex]\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}\\\\ [/tex]
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[tex]\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[/tex]
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[tex]x=\frac{140-10*5}{45}\\\\ x=\frac{140-50}{45}\\\\ x=\frac{90}{45}\\\\ x=2[/tex]
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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
[tex]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)[/tex]
[tex]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[/tex]