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 be the first term and D be the common difference of the AP. Therefore,

\[a_p = A + \left( p - 1 \right)D = a . . . . . \left( 1 \right)\]

\[ a_q = A + \left( q - 1 \right)D = b . . . . . \left( 2 \right)\]

\[ a_r = A + \left( r - 1 \right)D = c . . . . . \left( 3 \right)\]

Also, suppose A' be the first term and R be the common ratio of the GP. Therefore,

\[a_p = A' R^{p - 1} = a . . . . . \left( 4 \right)\]

\[ a_q = A' R^{q - 1} = b . . . . . \left( 5 \right)\]

\[ a_r = A' R^{r - 1} = c . . . . . \left( 6 \right)\]

Now,
Subtracting (2) from (1), we get

\[A + \left( p - 1 \right)D - A - \left( q - 1 \right)D = a - b\]

\[ \Rightarrow \left( p - q \right)D = a - b . . . . . \left( 7 \right)\]

Subtracting (3) from (2), we get

\[A + \left( q - 1 \right)D - A - \left( r - 1 \right)D = b - c\]

\[ \Rightarrow \left( q - r \right)D = b - c . . . . . \left( 8 \right)\]

Subtracting (1) from (3), we get

\[A + \left( r - 1 \right)D - A - \left( p - 1 \right)D = c - a\]

\[ \Rightarrow \left( r - p \right)D = c - a . . . . . \left( 9 \right)\]

\[\therefore a^{b - c} b^{c - a} c^{a - b}\]

` = [A'R ^((p-1))]^((q-r)D) xx [A'R^((q-1))]^((r-p)D) xx [A'R^((r-1))]^((p-q)D)  `  [Using (4), (5) (6), (7), (8) and (9)]

`= A'^((q-r)D) R^((p-1)(q-r)D)  xx A'^((r-p)D) R^((q-1)(r-p)D)  xx A'^((p-q)D) R^((r-1)(p-q)D)  `

`=A'^[[(q-r)D+(r-p)D+(p-q)D]] xx R^[[(p-1)(q-r)D+(q-1)(r-p)D+(r-1)(p-q)D]]`

`=A'^[[q-r+r-p+p-q]D] xx R^[[pq -pr - q+r+qr-pq -r +p+pr -qr -p+q]D]`

`= (A')^0 xx R^0`

`=1 xx 1`

`= 1`