Question

$1/(q+r);1/(r+p);1/(p+q)$ are in A.P., then

• A. $p,q,r$ are in A.P
• B. $p^2,q^2,r^2$ are in A.P
• C. $1/p,1/q,1/r$ are in A.P
• D. None of the above

Answer 1

Answer: (B)

$p^2,q^2,r^2$ are in A.P

$\frac{1}{(q+r)};\frac{1}{(r+p)};\frac{1}{(p+q)}$ are in A.P.
$\frac{2}{r+p}=\frac{1}{q+r}+\frac{1}{p+q}=\frac{p+2q+r}{(q+r)(p+q)}$
$\therefore 2(pq+q^2+rp+rq)=rp+2qr+r^2+p^2+2pq$
$\Rightarrow 2q^2=r^2+p^2$
$\therefore p^2,q^2,r^2$ are in A.P.