InterviewSolution
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In an H.P., pth term is qr and qth term is pr, show that rth term is pq. |
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Answer» Tp of H.P = qr ⇒ Tp of A.P. = \(\frac{1}{qr}\) ⇒ a + (p - 1)d = \(\frac{1}{qr}\) Where a and d are the first term and common difference respectively of the A.P Also Tq of H.P. = pr ⇒ Tq of A.P = \(\frac{1}{pr}\) ⇒ a + (q - 1)d = \(\frac{1}{pr}\) Eqn (ii) – Eqn (i) ⇒ (q – 1)d – (p – 1)d = \(\frac{1}{pr}\) - \(\frac{1}{qr}\) ⇒ (q – p)d = \(\frac{q-p}{pqr}\) ⇒ d = \(\frac{1}{pqr}\) ⇒ a + (p – 1)\(\frac{1}{pqr}\) = \(\frac{1}{qr}\) ⇒ a = \(\frac{1}{qr}\) - (p-1)\(\frac{1}{pqr}\) = \(\frac{1}{qr}\) - \(\frac{1}{qr}\) + \(\frac{1}{pqr}\) = \(\frac{1}{pqr}\) ∴ Tr of A.P = a + (r - 1)d = \(\frac{1}{pqr}\) + (r - 1)\(\frac{1}{pqr}\) = \(\frac{1}{pqr}\) + \(\frac{1}{pq}\) - \(\frac{1}{pqr}\) = \(\frac{1}{pq}\) ⇒ Tr of A.P = pq. |
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