CBSE Class 10 Answered

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

T_n = 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