Answer:
See Below.
Step-by-step explanation:
The first three terms of an A.P is equivalent to the first three terms of a G.P.
We want to show that this is only possible if r = 1 and d = 0.
If a is the initial term and d is the common difference, the A.P. will be:
[tex]a, a+d, \text{ and } a+2d[/tex]
Likewise, for the G.P., if a is the initial term (and it does not equal 0) and r is the common ratio, then our sequence is:
[tex]a, ar,\text{ and } ar^2[/tex]
The second and third terms must be equivalent. Thus:
[tex]a+d=ar\text{ and } a+2d=ar^2[/tex]
We can cancel the d. Multiply the first equation by -2:
[tex]-2a-2d=-2ar[/tex]
We can now add this to the second equation:
[tex](a+2d)+(-2a-2d)=(ar^2)+(-2ar)[/tex]
Simplify:
[tex]-a=ar^2-2ar[/tex]
Now, we can divide both sides by a (we can do this since a is not 0):
[tex]-1=r^2-2r[/tex]
So:
[tex]r^2-2r+1=0[/tex]
Factor:
[tex](r-1)^2=0[/tex]
Thus:
[tex]r=1[/tex]
The first equation tells us that:
[tex]a+d=ar[/tex]
Therefore:
[tex]a+d=a(1)\Rightarrow a+d=a[/tex]
Hence:
[tex]d=0[/tex]
Q.E.D.
