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A piece of equipment cost a certain factory 600,000 . If it depreciates in value $15 \%$ the first, $13.5 \%$ the next year, $12 \%$ the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
A piece of equipment cost a certain factory 600,000 . If it depreciates in value $15 \%$ the first, $13.5 \%$ the next year, $12 \%$ the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.7 | Maths | RD Sharma
A piece of equipment cost a certain factory 600,000 . If it depreciates in value $15 \%$ the first, $13.5 \%$ the next year, $12 \%$ the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Solution:

Given that a piece of equipment cost a certain factory is ₹ 600,000
We have to find the value of the equipment at the end of 10 years.
The price of equipment depreciates $15 \%, 13.5 \%, 12 \%$ in $1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}$ year and so on.
Therefore the A.P. will be $15,13.5,12, \ldots \ldots \ldots \ldots$ up to 10 terms
Here, $a=15, d=13.5-15=-1.5, n=10$
Using the formula,
$\begin{array}{l}
S_{n}=n / 2[2 a+(n-1) d] \\
S_{10}=10 / 2[2(15)+(10-1)(-1.5)] \\
=5[30+9(-1.5)] \\
=5[30-13.5] \\
=5[16.5] \\
=82.5
\end{array}$
The value of equipment at the end of 10 years is $=[100-$ Depreciation $\%] / 100 \times$ cost
$\begin{array}{l}
=[100-82.5] / 100 \times 600000 \\
=175 / 10 \times 6000 \\
=175 \times 600 \\
=105000
\end{array}$
As a result the value of equipment at the end of 10 years is ₹ 105000.

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