Question

suma a opt nr naturale impare consecutive este egala cu 192. Determinati cele opt nr .

Answer 1

(a+1)+(a+3)+(a+5)+(a+7)+(a+9)+(a+11)+(a+13)+(a+15)=192
8a+64=192
8a=128
a=128:8
a=16
deci numerele sunt
17,19,21,23,25,27,29,31
verificare

17+19+21+23+25+27+29+31=192
192=192

Answer 2

Daca n e par N+1+n+3+n+5+..n+15=192 8n+1+3+5+..+15=192 1+3+5+...+15=1+2+3+...+15-2-4-6-...-14=15*8-2*7*8/2=15*8-7*8=8*8=64 8n+64=192 Impartim prin 4 N+8=24 N=16 Nr sunt 17,19,21,23,25,27,29,31 Daca n impar N+n+2+n+4+n+6+n+8+n+10+n+12+n+14=192 8n+2+4+6+..+14=192 8n+2(1+2+3+...7)=192 8n+2*7*8/2=192 Impartim prin 8 N+7=24 N=17 Nr sunt 17 19 21 23 25 27 29 am calculat in 2 modalitati