Question

If 2(y - a) is the H.M. of y - x and y - z, then x - a, y - a, z - a are in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is B

GP

Let x - a = p, y - a = q, z - a = r

2(y - a) = 2q

y - x = (y - a) - (x - a) = q - p

y - z = (y - a) - (z - a) = q - r

As 2(y - a) is H.M. of y - x and y - z

2q is H.M. of q - p and q - r

$\Rightarrow$ 2q = $\frac{2(q-p)(q-r)}{q-p+q-r}$

q(2q - p - r) = (q - p)(q - r)

2 $q^2$-qp-qr= $q^2$-qr-pq+pr

$q^2$=pr

$\Rightarrow$ p, q, r are in G.P

Hence, (x - a), (y - a), (z - a) are in G.P