Arithmetic and Geometric Progression Question 1

$$a^2+4ad+4d^2=a^2+6ad+5d^2$$

$$2ad+d^2=0$$

$$d(2a+d)=0$$

$$d=0 \vee 2a+d=0$$

If $d$ is zero, then the ratio is clearly 1.

If $2a+d=0$ is zero, you have $d=-2a$. Substitute this in the first ratio we get $r=\frac{-3a}{-a}=3$.

$$a^2+4ad+4d^2=a^2+6ad+5d^2$$

This gives you $$d(2a+d)=0$$ so that either $d=0$ or $d= -2a$.

Your common ratio is given by $$\frac{a+2d}{a+d}$$ Substituting $d=0$ into that gives you the common ratio as $1$. This would make both the geometric and the arithmetic series constant, so we discard that solution.

Substitute $d=-2a$ into the common ratio equation to get $$r = 3.$$