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Question
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`
Solution
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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