a. Arithmetic progression: a1=k70,d=k3,n=120
S120=120⋅22a1+d(120−1)
=120⋅22(k70)+k3(120−1)=k29820 The cost of boring is k29820.
b. Let t= the time in months, C(t)= value of IPhone after t months
C(t)=32500−85t,t≥0If the Iphone’s value falls below k12500
C(t)<12500
32500−85t<12500
85t>32500−12500
85t>20000
t>8520000
t>235 It will take 236 months before the Iphone’s value falls below k12500.
c.
ui=5+2i,vi=2−5i
2ui−vi=2(5+2i)−(2−5i)=8−9i We know that
i=1∑n1=n
i=1∑ni=2n(n+1) Then
i=1∑5ui=i=1∑5(5+2i)=5i=1∑51+2i=1∑5i
=5(5)+2(25(5+1))=55
i=1∑5(2u1−vi)=i=1∑5(8+9i)=8i=1∑51+9i=1∑5i
=8(5)+9(25(5+1))=175
d. Let n= the week's number , An= the output of mobile phones per week.
e. Tecno
AnT=1000+200(n−1) Samsung
AnS=500(1.2)n−1
i)
A1T=1000+200(1−1)=1000
A5T=1000+200(5−1)=1800
A10T=1000+200(10−1)=2800
A15T=1000+200(15−1)=3800
A1S=500(1.2)1−1=500
A5S=500(1.2)5−1=1037
A10S=500(1.2)10−1=2580
A15S=500(1.2)15−1=6420
ii)
Tecno
n=1∑15(1000+200(n−1))=
=15(21000+3800)=36000 Samsung
n=1∑15(500(1.2)n−1)=
=500(1−1.21−(1.2)15)=36018
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this is brilliant, thank you so much for helping me
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