Zad1. W ciagu arytmetycznym a3=51 , a6=15 .Oblicz sume dziesieciu poczatkowych wyrazow tego ciagu ZAD2. W ciagu geometrycznym a2= -80, a5=10 .Oblicz sume siedmiu poczatkowych wyrazow tego ciagu
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
1
a3 = 51
a6 = 15
z własności ciągu:
a3 = a1+2r
a6 = a1+5r
a1+2r = 51
a1+5r = 15
-3r = 36
r = -12
a1-24 = 51
a1 = 75
a10 = a1+9r = 75 +9*-12 = 75 - 108 = -33
S10 = (75-33)/2 * 10 = 21*10 = 210
2
a2 = a1*q = -80
a5 = a1*q^4 = 10
a1 = -80/q
-80/q *q^4 = 10
-80q^3 = 10
q^3 = -1/8
q= -1/2
a1 = -80 : -1/2 = 160
S7 = 160 * (1-(-1/2^7)/1-1/2 = 160 * [1-(-1/128)]/1/2 = 160 * [129/128 * 2/1) = 160 * 129/64 = 20640/64 = 322,5