If \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) an

1.	If \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) and p, q, r are in A.P., show that x, y, z are in H.P.
Answer» Let \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) = k Then, \(\frac{a-x}{px}\) = k ⇒ \(\frac{a-x}{kx}\) = p ⇒ p = \(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\) Similarly, q = \(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\), r = \(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\) Now, p, q, r are in A.P \(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\) are in A.P. ⇒ \(\frac{a}{x}-1\), \(\frac{a}{y}-1\), \(\frac{a}{z}-1\) are in A.P. (Multiplying each term by k) ⇒ \(\frac{a}{x}\), \(\frac{a}{y}\), \(\frac{a}{z}\) are in A.P. (Adding 1 to each term) ⇒ \(\frac{1}{x}\), \(\frac{1}{y}\),\(\frac{1}{z}\) are in A.P. (Dividing each term by a) ⇒ x, y, z are in H.P

If \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) and p, q, r are in A.P., show that x, y, z are in H.P.

Answer»

Let \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) = k

Then, \(\frac{a-x}{px}\) = k ⇒ \(\frac{a-x}{kx}\) = p ⇒ p = \(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\)

Similarly, q = \(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\), r = \(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\)

Now, p, q, r are in A.P

\(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\) are in A.P.

⇒ \(\frac{a}{x}-1\), \(\frac{a}{y}-1\), \(\frac{a}{z}-1\) are in A.P. (Multiplying each term by k)

⇒ \(\frac{a}{x}\), \(\frac{a}{y}\), \(\frac{a}{z}\) are in A.P. (Adding 1 to each term)

⇒ \(\frac{1}{x}\), \(\frac{1}{y}\),\(\frac{1}{z}\) are in A.P. (Dividing each term by a)

⇒ x, y, z are in H.P