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Sa se det ratia unei progresii aritmetice (an)n>sau egal cu 1,a10-a2=16


Răspuns :

[tex] a_{10} - a_{2} =16\ \textless \ =\ \textgreater \ a_{1}+9r-( a_{1} +r)=16\ \textless \ =\ \textgreater \ a_{1}+9r-a_{1}-r=16[/tex]
Se reduc termenii asemenea , adica : a1 cu -a1
Ramane:
[tex]9r-r=16\ \textless \ =\ \textgreater \ 8r=16=\ \textgreater \ r= \frac{16}{8} =\ \textgreater \ r=2[/tex]
[tex]\displaystyle a_{10}-a_2=16 \\ a_{10-1}+r-(a_{2-1}+r)=16 \\ a_9+r-(a_1+r)=16 \\ \not a_1+9r-\not a_1-r=16 \\ 9r-r=16 \\ 8r=16 \\ r= \frac{16}{8} \\ r=2 [/tex]