Cho các số:$a_{1}$;$a_{2}$;$a_{3}$;…;$a_{100}$
$\frac{a1-1}{100}$=$\frac{a2-2}{99}$= $\frac{a3-3}{98}$=…=$\frac{a100-100}{1}$
Và $a_{1}$+$a_{2}$+$a_{3}$+…+$a_{100}$ = 10100
Cho các số:$a_{1}$;$a_{2}$;$a_{3}$;…;$a_{100}$ $\frac{a1-1}{100}$=$\frac{a2-2}{99}$= $\frac{a3-3}{98}$=…=$\frac{a100-100}{1}$ Và $a_{1}$+$a_{2}$+
By Clara
`(a_1-1)/100=(a_2-2)/99=(a_3-3)/98=…=(a_100-100)/1`
Áp dụng t/c dãy tỉ số = nhau:
`(a_1-1+a_2-2+a_3-3+…+a_(100)-100)/(100+99+…+1)`
`=((a_1+a_2+…+a_(100))-(1+2+…+100))/(100+99+…+1)`
`=(a_1+a_2+…+a_(100))/(100+99+…+1) -1`
`=10100/5050 -1=2-1=1`
Suy ra:
`(a_1-1)/100=1 => a_1-1=100 => a_1=101`
…v.v
`(a_100-100)/1=1 => a_100-100=1 => a_100=101`
`⇒a_1=a_2=…=a_(100)=101`
Đáp án:
`(a_1-1)/100=(a_2-2)/99=…=(a_100-100)/1`
`=(a_1-1+a_2-2+…+a_100-100)/(100+99+…+1)`
`=((a_1+a_2+…+a_100)-(1+2+…+100))/(5050)`
`=(10100-5050)/(5050)`
`=5050/5050=1`
`=>`
`(a_1-1)/100=1 => a_1-1=100 => a_1=101`
`(a_2-2)/99=1 => a_2-2=99 => a_2=101`
`…`
`(a_100-100)/1=1 => a_100-100=1 => a_100=101`
Vậy `a_1=a_2=a_3=…=a_100=101`