Question

Fie sirul:4,10,16,22,28... a) Copletati sirul cu inca doi termeni b)Aflati T 2003 c) Calculati suma primilor 23 de termeni d) Verificati daca 202 este termen al sirului

Answer 1

Se observa cea elementele cresc din 6 in 6:

a)
4, 10, 16, 22, 28, 34, 40, ...

b)
T₁ = 4
T₂ = 10 = T₁ + 6
T₃ = 16 = T₂ + 6 = T₁ + 6 + 6 = T₁ + 2 * 6
T₄ = 22 = T₃ + 6 = ... = T₁ + 6 + 6 + 6 = T₁ + 3 * 6
T₅ = 28 = T₄ + 6 = T₁ + 4 * 6
T₆ = 34 = T₁ + 5 * 6

Se observa ca Tn = T₁ + (n - 1)  * 6

T₂₀₀₃ = T₁ + 2002 * 6 = 4 + 2002 * 6 = 12016

c)
[tex]S=T_1+T_2+T_3+...+T_{23}\\ T_n=T_1+6(n-1) \\S=(T_1+0\cdot6)+(T_1+1\cdot6)+(T_1+2\cdot6)+...+(T_1+22\cdot6)\\ S=\underbrace{T_1+T_1+...+T_1}_{\text{23 de elemente}}+6+2\cdot6+...+22\cdot6\\\\ S=23T_1+6(1+2+...+22)\\ S=23\cdot4+6\cdot\frac{22\cdot23}{2}=23\cdot4+1518=\boxed{1610}[/tex]

d)
Presupunem ca 202 este termen al sirului, asadar exista un n pentru:
[tex]T_n=T_1+6(n-1)=202\\ 4+6(n-1)=202\\ 6(n-1)=198\\ n-1=\frac{198}{6}=33\\ \boxed{n=34}[/tex]

Asadar, T₃₄ = 202, deci 202 este termen al sirului