Question

Intr-o progresie aritmetica, in care Sn este suma primilor n termeni, avem S1000=S2000. Sa se calculeze S3000.

Answer 1

Răspuns:

Explicație pas cu pas:

$\\ S_{n} =\frac{(a_{1}+a_{n)}  n}{2} \\ \\ S_{1000} =\frac{(a_{1}+a_{1000)}  1000}{2} , S_{2000} =\frac{(a_{1}+a_{2000)} 2000}{2} \\ \\ \frac{(a_{1}+a_{1000)}  1000}{2}=\frac{(a_{1}+a_{2000)}  2000}{2}\\ \\ a_{1} +a_{1000} = 2a_{1} +2a_{2\\ 000} \\ \\ a_{1}+a_{1}+999r=2a_{1}+2a_{1}+3998r\\ \\ 2a_{1}=-2999r,2a_{1}+2999r=0 \\ \\ S_{3000} =\frac{(a_{1}+a_{3000)} 3000}{2}=\frac{(2a_{1}+2999r)3000} 2=\frac{0x3000}{2} =\frac{0}{2}=0$