104 MORE INTERESTFORMULAS
gradient. Instead, we willsubtract an increasing gradient from an assumed uniform series of
payments.
6000
o t
0-1-2-3-4 - 0-1-2-3-4
24'
[
000 18,000
i
12,000
t 60;0
= 0-1-2-3-4
1 1 1 1
A' A' A' A'
A'= 24,000- 6000(AjG,10%,4)
=,24,000 -- 6000(1.38t)
= 15,714 rupees
The projected equivalent uniform maintenanceand repair cost is 15,714rupees per year.
oS
Computethe value ofPin the diagram. Use a 10% interest rate.
150
50
o t
0-1-2-3-4-5~6
100
r
p
~.,,_. "'.-; ",
SptlJ11()~' ,.
With the arithmetic gradient series present worth factor, we can compute a presentsumJ. I
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