Question

Rezolvati in R ecuatia 5+11+17+....+x=320​

Answer 1

5 + 11+17+...+x = 320

x + (x-6)+(x-12)+...+5 = 320

(x+5)+(x+5)+(x+5)+...+(x+5) = 320•2

n(x+5) = 320•2

5+11+17+...+x = 320

(6•1-1)+(6•2-2)+....+(6•(x+1)/6-1) = 320

=> n = (x+1)/6

=> (x+1)/6 • (x+5) = 320•2

=> (x+1)(x+5) = 320•12 = 32•120 = 64•60

=> (x+1)(x+5) = (59+1)•(59+5)

=> x = 59