Question 1072878
<pre><font size=5><b>
When something depreciates and 15%, its value
is multiplied by 100%-15% or 85% or 0.85.

So the geometric sequence is 

After     
 this      The
Number    value    
  of       of
years     the     
have      lathe
passed     is 
----------------------
  0       £3000
  1       £2550 
  2       £2167.50
  3       £1842.375
  4       £1566.01875
  5       £1551.115938
  6       £1121.448547
  7       £ 961.7312648
  8       £ 817.4715751
  9       £ 694.8508388
 10       £ 590.623213
 11       £ 502.0297311

The lathe will be sold after 11 years.

That's the easiest way, but your teacher perhaps
wants you do it using the formula for the nth term
of a geometric series:

By formula, a<sub>1</sub> = £3000, r = 0.85

{{{a[1]*r^(n-1)}}}{{{""<""}}}{{{550}}}
{{{3000*(0.85)^(n-1)}}}{{{""<""}}}{{{550}}}
{{{(0.85)^(n-1)}}}{{{""<""}}}{{{550/3000}}}

Take logarithms of both sides (you may use
either base logarithms, natural or common))

{{{log((0.85)^(n-1)))}}}{{{""<""}}}{{{log((550/3000))}}}

{{{(n-1)log((0.85))}}}{{{""<""}}}{{{log((11/60))}}}

Since log(0.85) is negative, we must reverse
the inequality symbol from < to > when we
divide both sides by log(0.85)

{{{n-1}}}{{{"">""}}}{{{log((11/60))/log((0.85))}}}

{{{n-1}}}{{{"">""}}}{{{10.5401789}}}

{{{n}}}{{{"">""}}}{{{11.5401789}}}
 
The lathe will be sold after 11 years.

Edwin</pre></b></font>