Question

If a,b,c are in AP or GP or HP, then $\frac{a-b}{b-c}$ is equal to

• A. $\frac{b}{a}$ or $1$ or $\frac{b}{c}$
• B. $\frac{c}{a}$ or $\frac{c}{b}or1$
• C. $1$ or $\frac{a}{b}$ or $\frac{a}{c}$
• D. $1$ or $\frac{a}{b}$ or $\frac{c}{a}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{c}{a}$ or $\frac{c}{b}or1$
B $\frac{b}{a}$ or $1$ or $\frac{b}{c}$
C $1$ or $\frac{a}{b}$ or $\frac{a}{c}$
D $1$ or $\frac{a}{b}$ or $\frac{c}{a}$

If $a$, $b$ and $c$ are in AP, then common difference $d=c-b=b-a$

$\therefore \frac{a-b}{b-c}=1$

Also, $b=a+d$ and $c=b+d$

If $a$, $b$ and $c$ are in GP let the common ratio be $r$.

Then $b=ar$ and $c=br$.

$\frac{a-b}{b-c}=\frac{a-ar}{b-br}=\frac{a}{b}$

Also $c=a{r}^2$

$\frac{a-b}{b-c}=\frac{a-ar}{ar-a{r}^2}=\frac{1}{r}=\frac{ar}{a{r}^2}=\frac{b}{c}$

So, $\frac{a-b}{b-c}=\frac{a}{b}=\frac{b}{c}$

If $a,b,c$ are in HP, then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP.

Then, $b=\frac{2ac}{a+c}$

$\frac{a-b}{b-c}=\frac{a-\frac{2ac}{a+c}}{\frac{2ac}{a+c}-c}=\frac{a^2-ac}{ac-c^2}=\frac{a(a-c)}{c(a-c)}=\frac{a}{c}$