Question 974313
What is the first 50 terms of a sequence whose 53th and 55th term are both 158?
<pre>
If it is an arithmetic sequence then 

{{{a[53]+d=a[54]}}} and {{{a[54]+d=a[55]}}}

{{{158+d=a[54]}}} and {{{a[54]+d=158}}
                     {{{a[54]=158-d}}}

158+d = 158-d  since both equal a<sub>54</sub>
   2d = 0
    d = 0

Therefore all the terms are 158, that is, the sequence is

158, 158, 158, 158, 158, ...

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If it is a geometric sequence then 

{{{a[53]r=a[54]}}} and {{{a[54]r=a[55]}}}

{{{158r=a[54]}}} and {{{a[54]r=158}}}
                     {{{a[54]=158/r}}}

{{{158r = 158/r}}}  since both equal a<sub>54</sub>
{{{158r^2=158}}}
{{{r^2=1}}}
{{{r="" +- 1}}}

If r = 1, then all the terms are 158  
    
If r = -1, then the terms alternate between 158 and -158. With the
odd numbered terms being 158 and the even numbered terms being -158

So either way the sequence is either:

158, 158, 158, 158, 158, ...

or

158, -158, 158, -158, 158, -158, ...

Edwin</pre>