Question 1197301
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Let's use the standard notation of a for the first term and r for the common ratio.  Then<br>
The sum of the first and fourth terms is 195:
{{{a+ar^3=195}}}
{{{a(1+r^3)=195}}}  [1]<br>
The sum of the second and third terms is 60:
{{{ar+ar^2=60}}}
{{{a(r)(1+r)=60}}} [2]<br>
It should be clear that both a and r are integers, since both sums are integers.<br>
So a quick informal method of solving the problem is to look at equation [1] and find what integer r can be so that (1+r^3) is a factor of 195.<br>
r=1 --> 1+r^3=2 --> not a factor of 195
r=2 --> 1+r^3=9 --> not a factor of 195
r=3 --> 1+r^3=28 --> not a factor of 195
r=4 --> 1+r^3=65 --> YES -- 195 = 3(65)<br>
So it looks as if a=3 and r=4 are the values we are looking for.  Let's check that.<br>
The sequence is 3, 12, 48, 192.  3+192 = 195, and 12+48 = 60.<br>
ANSWER: The geometric sequence is 3, 12, 48, and 192.<br>
Note that the sequence could also be 192, 48, 12, 3 -- with first term 192 and common ratio 1/4.<br>
For a formal (and ugly) algebraic solution, using equations [1] and [2] above....<br>
{{{(a(1+r^3))/(a(r)(1+r))=195/60=13/4}}}
{{{(a(1+r)(1-r+r^2))/(a(r)(1+r))=195/60}}}
{{{(1-r+r^2)/r=13/4}}}
{{{4(1-r+r^2)=13r}}}
{{{4-4r+4r^2=13r}}}
{{{4r^2-17r+4=0}}}
{{{(4r-1)(r-4)=0}}}<br>
This gives us the two values we found above for the common ratio: r=4 or r=1/4.<br>
And that leads us quickly to the two sequences we found.<br>