If p^th, q^th, and r^th terms of an A.P. and G.P. are both a, b and c respectively, show that

a^b–c . b^{c – a} . c^{a – b} = 1

Let the first term of AP be m and common difference as d

Let the GP first term as l and common ratio as s

The n^th term of an AP is given as t_n = a + (n – 1)d where a is the first term and d is the common difference

The n^th term of a GP is given by t_n = ar^n-1 where a is the first term and r is the common ratio

The p^th term (t_p) of both AP and GP is a

For AP

⇒ t_p = m + (p – 1)d

⇒ a = m + (p – 1)d …(u)

For GP

⇒ t_p = ls^p-1

⇒ a = ls^p-1 …(v)

The q^th term (t_q) of both AP and GP is b

For AP

⇒ t_q = m + (q – 1)d

⇒ b = m + (q – 1)d …(w)

For GP

⇒ t_q = ls^q-1

⇒ b = ls^q-1 …(x)

The r^th term (t_r) of both AP and GP is c

For AP

⇒ t_r = m + (r – 1)d

⇒ c = m + (r – 1)d …(y)

For GP

⇒ t_r = ls^r-1

⇒ c = ls^r-1 …(z)

Let us find b – c, c – a and a – b

Using (w) and (y)

⇒ b – c = (q – r)d …(i)

Using (y) and (u)

⇒ c – a = (r – p)d …(ii)

Using (u) and (w)

⇒ a – b = (p – q)d …(iii)

We have to prove that a^b–c.b^c–a.c^a–b = 1

LHS = a^b–c.b^c–a.c^a–b)

Using (v), (x) and (z)

⇒ LHS = (ls^p-1)^b-c.(ls^q-1)^c-a.(ls^r-1)^a-b

= s^p(b-c).s^q(c-a).s^r(a-b)

Substituting values of a – b, c – a and b – c from (iii), (ii) and (i)

= s^p(q-r)d.s^q(r-p)d.s^r(p-q)d

= s^pqd-prd.s^qrd-pqd.s^prd-qrd

= s^{pqd-prd+qrd-pqd+prd-qrd}

= s⁰ = 1

⇒ LHS = RHS

Hence proved