Question

Fie $(a_{n})_{n\geq1}$ o progresie aritmetica, cu $a_{1}$=-4 so r=$\frac{1}{3}$. Cati termeni subunitari are progresia?

Answer 1

-4+(5•3)/3 = -4+15•(1/3) = -4+5 = 1

=> Avem 1+14 = 15 termeni subunitari.

Answer 2

$\it a_n=a_1+(n-1)r \Rightarrow a_n=-4+(n-1)\cdot\dfrac{1}{3} =\dfrac{n-13}{3}\\ \\ a_n<1 \Rightarrow \dfrac{n-13}{3}<1 \Rightarrow n-13<3|_{+13} \Rightarrow n<16 \Rightarrow n\leq 15\\ \\ Deci,\ progresia\ are 15\ termeni\ subunitari.$