Question

If p, q, r are in AP as well as GP, then which one of the following is correct ?

• A. $p=q\ne r$
• B. $p\ne q\ne r$
• C. $p\ne q=r$
• D. $p=q=r$

Answer 1

The correct option is D $p=q=r$

For A.P

$q=\frac{p+r}{2}$

For G.P

$q^2=pr$

Now combining both equations, we get

$\frac{(p+r{)}^2}{4}=pr$

$\Rightarrow (p+r{)}^2=4pr$

$\Rightarrow p^2+r^2+2pr=4pr$

$\Rightarrow p^2+r^2-2pr=0$

$\Rightarrow (p-r{)}^2=0$

$\Rightarrow p=r$

Using this in equation $q=\frac{p+r}{2}$ we have $p=q=r$

Hence, option D is correct.