Question

1.oblicz sume 1+3+5+7+...201

bo muszę sprawdzić czy mam dobrze


zad2.
oblicz wyraz pierwszy i róznice r
1.a7-a3=8 i a2*a7=75
2.a5=2 i a100=97
3.a2+a5=8 i a3+a7=17

Answer 1

Z.1
1 +3 + 5 +7 + ... + 201
Jest to ciąg arytmetyczny:
a1 = 1
an = 201
r = 2
an = a1 + (n-1)*r
201 = 1 +(n -1)*2 = 1 +2n -2 = 2n -1
2n = 201 +1 = 202
n = 202 :2 = 101
S101 = {[a1 + a101]/2}*101 ={[1 + 201]/2}* 101 =[202/2]*101 =
= 101 * 101 = 10 201
z.2
a)
a7 - a3 = 8 oraz a2 * a7 = 75
zatem
(a1 +6r) -(a1 +2r) = 6r - 2r = 4r = 8 ---> r = 8 : 4 = 2
r = 2
oraz (a1 +r)*(a1 + 6r) = (a1 +2)*(a1 + 6*2) = 75
czyli
(a1 +2)*(a1 + 12) = 75
(a1)² + 2 a1 + 12 a1 + 24 = 75
(a1)² + 14 a1 - 51 = 0
Δ = 14² - 4*1*(-51) = 196 + 204 = 400
√Δ = 20
zatem
a1 = [-14 - 20]/2 = -17 lub a1 = [ -14 + 20]/2 = 3
Odp. a1 = -17 oraz r = 2
lub
a1 = 3 oraz r = 2.
b)
a5 = 2 oraz a100 = 97
zatem
a1 + 4r = 2
a1 +99r = 97
(a1 +99r) -( a1 + 4r) = 95r = 97 - 2 = 95
95r = 95 ---> r = 1
a1 = 2 - 4r = 2 - 4 * 1 = 2 - 4 = -2
Odp. a1 = -2 oraz r = 1
c)
a2 + a5 = 8 oraz a3 + a7 = 17
(a1 +r) +(a1 + 4r) = 8
(a1 + 2r) +(a1 + 6r) = 17
-----------------------------
2 a1 + 5r = 8
2 a1 + 8r = 17
--------------------
(2a1 + 8r) - ( 2 a1 + 5r) = 8r - 5r = 3r = 17 - 8 = 9
3r = 9 ---> r = 9 : 3 = 3
r = 3
2 a1 = 8 - 5r = 8 - 5*3 = 8 - 15 = - 7
a1 = - 7 : 2 = -3,5
Odp. a1 = -3,5 oraz r = 3
=====================================