arithmetic progression 1

Asked by  | 2nd May, 2009, 08:11: AM

Expert Answer:

The cost of the equipment after the end of first year = 600000 - (15% of 600000) = 510000

The cost of equipment after the end of secon year = 600000 - (13.5% 0f 600000) = 519000

The cost of the equiment after the end of third year = 600000 - (12% of 600000) = 528000

Here we observe that after a 15% depreciation in the cost, the cost of the equipment increases by 1.5% (9000) every year. So, we can form an AP whose first term is 510000 and common difference is 9000

The AP is 510000,519000,528000..........

a = 510000

d=9000

By using the standard formula of nth term of an AP

Tn = a+(n-1)d = 510000 +(10-1) 9000

                         = 510000 +81000

                         = 591000

The cost of the equiment at the end of 10 years would be 591000

Answered by  | 4th May, 2009, 02:52: PM

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