Question 1203113
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An employee starts work on 1 January 2000 on an annual salary of $30,000. 
His pay scale will give him an increase of $800 per annum 
on the first of January until 1 January 2015 inclusive. 
He remains on this salary until he retires on 31st December 2040.
How much will he earn during his working life.
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<pre>
His annual salaries from 01/01/2000 to 12/31/2015 form an arithmetic progression
with the first term {{{a[2000]}}} = 30000 dollars and common difference d= 800 dollars.


In particular, his annual salary for the year 2015 is 

    {{{a[2015]}}} = 30000 + 15*800 = 42000 dollars  (the 16-th term of the AP).


Thus, from 01/01/2000 till 12/31/2015 he earned  {{{((30000 + 42000)/2)*16}}} = 576000 dollars.


From 01/01/2016 to 12/31/2040, his annual earning rate was constant 42000 dollars per year 
during  (40-15) = 25 years.

So, during these 25 years, he earns  25*42000 = 1050000 dollars.


The total earning during his working life will be this sum

    576000 + 1050000 = 1,626,000 dollars.    <U>ANSWER</U>
</pre>

Solved.


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For introductory lessons on arithmetic progressions see 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/The-proofs-of-the-formulas-for-arithmetic-progressions.lesson>The proofs of the formulas for arithmetic progressions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Problems-on-arithmetic-progressions.lesson>Problems on arithmetic progressions</A>  

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Word-problems-on-arithmetic-progressions.lesson>Word problems on arithmetic progressions</A>

in this site.