If pth, qth and rth terms of an A.P. and G.P. are both a, b and c resp

1.	If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab – cb c – aca – b = 1.
Answer» Let the A.P. be A, A + D, A + 2 D, ... and G.P be x, xR, xR², ... then a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D ⇒ a – b = (p – q)D Also, b – c = (q – r)D And, c – a = (r – p)D Also a = p^th term of GP ∴ a = xR^{p – 1} Similarly, b = xR^{q – 1} & c = xR^{r – 1} Hence, (a^{b – c} ).(b^{c – a} ).(c^{a – b}) = [(xR^{p – 1}) ^{(q – r)D}].[(xR ^{q – 1}) ^{(r – p)D}].[(xR ^{r – 1) (p – q)D}] = x ^{(q – r + r – p + p – q)D}. R ^{[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]D} ⇒ (a^b ^{– c} ).(b^c ^{– a} ).(c ^{a – b}) = x⁰. R⁰ ⇒ (a ^{b – c} ).(b ^{c – a} ).(c ^{a – b}) = 1 …proved

If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab – cb c – aca – b = 1.

Answer»

Let the A.P. be A, A + D, A + 2 D, ... and G.P be x, xR, xR², ... then

a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D

⇒ a – b = (p – q)D

Also, b – c = (q – r)D

And, c – a = (r – p)D

Also a = p^th term of GP

∴ a = xR^{p – 1}

Similarly, b = xR^{q – 1} & c = xR^{r – 1}

Hence,

(a^{b – c} ).(b^{c – a} ).(c^{a – b}) = [(xR^{p – 1}) ^{(q – r)D}].[(xR ^{q – 1}) ^{(r – p)D}].[(xR ^{r – 1) (p – q)D}]

= x ^{(q – r + r – p + p – q)D}. R ^{[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]D}

⇒ (a^b ^{– c} ).(b^c ^{– a} ).(c ^{a – b}) = x⁰. R⁰

⇒ (a ^{b – c} ).(b ^{c – a} ).(c ^{a – b}) = 1 …proved