Question

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

Answer 1

Let A and d be the first term and common difference respectively of an A.P. and x and R be the first term and common ratio respectively of the G.P.

∴ A + (p – 1)d = a   .....(i)

A + (q – 1)d = b   .....(ii)

And A + (r – 1)d = c   ......(iii)

For G.P., we have

xR^p–1 = a  .....(iv)

xR^q–1 = b  .....(v)

And xR^r–1 = c   .....(vi)

Subtracting equation (ii) from equation (i) we get

(p – q)d = a – b   ......(vii)

Similarly, (q – r)d = b – c   ......(viii)

And (r – p)d = c – a   ......(ix)

Now we have to prove that

a^b–c . b^c–a . c^a–b = 1

L.H.S. a^b–c . b^c–a . c^a–b

= `[x"R"^(p - 1)]^((q - r)d) * [x"R"^(q - 1)]^((r - p)d) * [x"R"^(r - 1)]^((p - q)d)`  ....[From (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix)]

= `x^((q - r)d) * "R"^((p - 1) (q - r)d) * x^((r - p)d) * "R"^((q - 1) (r - p)d) * x^((p - q)d) * "R"^((r - 1)(p - q)d)`

= `x^((q - r)d + (r - p)d) "R"^((p - 1)(q - r)d + (q - 1)(r - p)d + (r - 1)(p - q)d)`

= `x^((q-r + r - p + p - q)d) * "R"^((pq - pr - q + r + qr - pq - r + p + pr + pr - qr - p + q)d)`

= `x^((0)d) * "R"^((0)d)`

= `x^0 * "R"^0`

= 1 R.H.S.

L.H.S. = R.H.S. 

Hence proved.