Question

escreva a P.A em que o 4° termo  vale 24 e o 9°termo vale 49

Answer 1

$a_{n}=a_{1}+(n-1)r\\a_{4}=a_{1}+(4-1)r\\a_{4}=a_{1}+3r\\\\a_{9}=a_{1}+(9-1)r\\a_{9}=a_{1}+8r$
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$a_{4}=24\\a_{1}+3r=24$

$a_{9}=49\\a_{1}+8r=49$
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Sistema:

$\left \{ {{a_{1}+3r=24} \atop {a_{1}+8r=49}} \right.$

Subtraindo uma pela outra:

$a_{1}-a_{1}+3r-8r=24-49\\-5r=-25\\5r=25\\r=25/5\\r=5$

$a_{1}+3r=24\\a_{1}+3*5=24\\a_{1}+15=24\\a_{1}=24-15\\a_{1}=9$
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P.A:

$a_{1}=9\\a_{2}=a_{1}+r=9+5=14\\a_{3}=a_{2}+r=14+5=19\\a_{4}=a_{3}+r=19+5=24\\a_{5}=a_{4}+r=24+5=29\\a_{6}=a_{5}+r=29+5=34\\a_{7}=a_{6}+r=34+5=39\\a_{8}=a_{7}+r=39+5=44\\a_{9}=a_{8}+r=44+5=49$