Question

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

Answer 1

$a_{10} - a_{2} =16\ \textless \ =\ \textgreater \ a_{1}+9r-( a_{1} +r)=16\ \textless \ =\ \textgreater \ a_{1}+9r-a_{1}-r=16$
Se reduc termenii asemenea , adica : a1 cu -a1
Ramane:
$9r-r=16\ \textless \ =\ \textgreater \ 8r=16=\ \textgreater \ r= \frac{16}{8} =\ \textgreater \ r=2$

Answer 2

$\displaystyle a_{10}-a_2=16 \\ a_{10-1}+r-(a_{2-1}+r)=16 \\ a_9+r-(a_1+r)=16 \\ \not a_1+9r-\not a_1-r=16 \\ 9r-r=16 \\ 8r=16 \\ r= \frac{16}{8} \\ r=2$