Question

afla suma a zece nr consecutive s
tiind ca primul nr e de 4 ori mai mic decat ultimul

Answer 1

Notam cele 10 numere consecutive cu a1,a2,a3,a4,...,a10.
[tex]a1= \frac{a10}{4} \\ a10=a1+9 \\ a1= \frac{a1+9}{4} \\ 4a1=a1+9 \\ 3a1=9 \\ a1=3 \\ Numere-consecutive : a2=a1+1=4 \\ a3=a1+2=5 \\ a4=a1+3=6 \\ a5=a1+4=7 \\ a6=a1+5=8 \\ a7=a1+6=9 \\ a8=a1+7=10 \\ a9=a1+8=11 \\ a10=a1+9=12 \\ S=a1+a2+...+a10= \frac{(a10+a1)*10}{2} = \frac{(12+3)*10}{2} = \frac{15*10}{2} =75 [/tex]

Answer 2

a+(a+1)+(a+2)+(a+3)+...+(a+9)=?
10a+45=?
a+9=4a
3a=9
a=3
10a+45=10*3+45=30+45=75