Question

Sa se rezolve ecuatiile:
a) 2+7+12+...+x=1550
b) 3+6+9+...+x=2460
c) 4+10+16+...+x= 3104
d) (x+1)+(x+9)+(x+17)+...+(x+393)=10100
Stiu doar ca sunt progresii aritmetice... Nu le-am inteles foarte bine, as dori un model de rezolvare pentru acest tip de exercitiu sa-l stiu odata pentru totdeauna, va rog frumos... daca aveti rabdare... Multumesc frumos! Ofer 54 puncte!

Answer 1

a)
2+7+12+...+x=1550
2+2+5+2+10+....+x=1550
nr de termeni de 2
(x-2):5+1=x/5-2/5+1=x/5+3/5=(x+3)/5
nr de termeni multiplu de 5
(x+3)/5-1=(x+3-5)/5=(x-2)/5
2+2+5+2+10+....+x=1550
2*(x+3)/5+5(1+2+3+...+(x-2)/5)=1550
(2x+6)/5+5*(x-2)/5*[(x-2)/5+1]/2=1550
(2x+6)/5+(x-2)*(x-2+5)/10=1550
(2x+6)/5+(x-2)(x+3)/10=1550
4x+12+x²-2x+3x-6=15500
x²+5x-15494=0
Δ=5²+4×15494=61976
x=(-5+√61976)/2=(-5+249)/2=122

verificare
2+7+12+...122=
nr de termeni de 2= (x+3)/5=(122+3)=125/5=25
nr de termeni multiplu de 5=(x-2)/25=120/5=24
=2+2+5+2+10+2+....+2+120=
=2*25+5*24*25/2=50+1500=1550


daca ai invatat progresii aritmetice
Sn={[a₁+a(n)]*n}/2
Sn=1550
a₁=2
a(n)=x
n=(x-2):5+1=x/5-2/5+1=x/5+3/5=(x+3)/5
1550=[(2+x)*(x+3)/5]/2
3100=(2x+x²+6+3x)/5
15500=x²+5x+6
x²+5x+6=15494
Δ=5²+4×15494=61976
x=(-5+√61976)/2=(-5+249)/2=122