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Question
If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that \[a^{b - c} b^{c - a} c^{a - b} = 1\]
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Solution
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`
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