Question

The 1st 3 terms of an A.P., a, a+d, and a+2d, are the same as the 1st 3 terms of a G.P.( a is not equal to 0). Show that this is only possible if r=1 and d=0

Answer 1

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:

$a, a+d, \text{ and } a+2d$

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:

$a, ar,\text{ and } ar^2$

The second and third terms must be equivalent. Thus:

$a+d=ar\text{ and } a+2d=ar^2$

We can cancel the d. Multiply the first equation by -2:

$-2a-2d=-2ar$

We can now add this to the second equation:

$(a+2d)+(-2a-2d)=(ar^2)+(-2ar)$

Simplify:

$-a=ar^2-2ar$

Now, we can divide both sides by a (we can do this since a is not 0):

$-1=r^2-2r$

So:

$r^2-2r+1=0$

Factor:

$(r-1)^2=0$

Thus:

$r=1$

The first equation tells us that:

$a+d=ar$

Therefore:

$a+d=a(1)\Rightarrow a+d=a$

Hence:

$d=0$

Q.E.D.