Question

Let P, Q , R be defined as
$P=a^2+a{b}^2-a^{2}c-a{c}^2$
$Q=b^2+b{c}^2-a^{2}b-a{b}^2$
$R=a^{2}c+c^{2}a-c^{2}b-c{b}^2$
where a, b, c are all +ive and the equation $P{x}^2+Qx+R=0$ has equal roots then a , b , c are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

Answer: (C)

H.P.

$P=a^2(b-c)+a(b^2-c^2)=a(b-c)[a+b+c]$
$\therefore P+Q+R=(a+b+c)\sum a(b-c)=0$
Hence $P{x}^2+Qx+R=0$
Since it has equal roots therefore roots are 1,1
Product of roots 1,1 = 1 = $\frac{R}{P}$
or P = R or (a + b + c) a(b - c) = (a + b + c) c(a - b)
or a(b - c) = c(a - b)
or b(a + c) =2ac or b =$\frac{2ac}{a+c}$
or a,b,c, are in H.P.