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If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`.
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Let the first term of the geometric progression be A and the common ratio be R.
pth term = ARp – 1 = a ...(i)
qth term = ARq – 1 = b ...(ii)
rth term = ARr – 1 = c ...(iii)
Using q – r of equation (i), r – p of equation (ii), p – q power of equation (iii),
aq−r. br−p. cp−q = (ARp−1)q −r. (ARq−1)r−p. (ARr−1)p−q
= `A^(q - r + r - p + p - q) R^((p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q))`
= `A^0. R^(p (q - r) - 1 (q - r) + q (r - p) - 1(r - p) + r (p - q) - 1(p - q))`
= `R^(pq - pr - q + r + qr- pq - r + p + rp - rp - p + q)`
= R0
= 1
Thus, the given result is proved.
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