104 MORE INTERESTFORMULASgradient. Instead, we willsubtract an increasing gradient from an assumed uniform series of
payments.6000
o t
0-1-2-3-4 - 0-1-2-3-424'[
000 18,000i
12,000t 60;0
= 0-1-2-3-41 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.
oSComputethe value ofPin the diagram. Use a 10% interest rate.
15050o t
0-1-2-3-4-5~6100r
p~.,,_. "'.-; ",SptlJ11()~' ,.
With the arithmetic gradient series present worth factor, we can compute a presentsumJ. I
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