(a) \(-\frac{q}{2}\)
p, q, r are in H.P. ⇒ q = \(\frac{2pr}{p+r}\) .....(i)
Let a and d be the first term and common difference respectively of the A.P.
∴ Tp + 1 = a + pd
Tq + 1 = a + qd
Tr + 1 = a + rd
Given, Tp + 1, Tq + 1, Tr + 1 are in G.P.
⇒ (Tq + 1)2 = Tp + 1 · Tr + 1
(a + qd)2 = (a + pd) (a + rd)
⇒ a2 + 2aqd + q2d2 = a2 + ad (p + r) + prd2
⇒ ad (2q – p – r) = d2 (pr – q2)
⇒ \(\frac{a}{d}\) = \(\frac{pr-q^2}{2q-p-r}\) = \(\frac{\frac{1}{2}(p+r)q-q^2}{2q-(p+r)}\) (Using (i))
= \(\frac{-\frac{q}{2}[2q-(p+r)]}{2q-(p+r)}\) = \(-\frac{q}{2}.\)