Question

Fie $a_{n}$ ,n≥1 o progresie aritmetica. Stiind ca $a_{3}+ a_{19}=10$ , sa se calculeze $a_{6}+ a_{16}$.

Answer 1

[tex]a_{6}=a_{3} + 3r\\ a_{16}=a_{19} - 3r \\ a_{6}+a_{16}= a_{3} + 3r + a_{19} - 3r = a_3 + a_{19} = 10[/tex]