Example 22 - If a, b, c are in GP and a1/x = b1/y = c1/z - AP and GP mix questions

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Example 22 If a, b, c are in G.P. and "a" ^(1/๐‘ฅ) = "b" ^(1/๐‘ฆ) = "c" ^(1/๐‘ง) , prove that x, y, z are in A.P. Given that "a" ^(1/๐‘ฅ) = "b" ^(1/๐‘ฆ) = "c" ^(1/๐‘ง) Let "a" ^(1/๐‘ฅ) = "b" ^(1/๐‘ฆ) = "c" ^(1/๐‘ง) = k Now, "a" ^(1/๐‘ฅ) = k Taking power x both sides ("a" ^(1/๐‘ฅ) )^๐‘ฅ = ใ€–"(k)" ใ€—^๐‘ฅ "a" ^(๐‘ฅ ร— 1/๐‘ฅ) = "k" ^๐‘ฅ a = "k" ^๐‘ฅ Also, "b" ^(1/๐‘ฆ) = k Taking power y both sides ("b" ^(1/๐‘ฆ) )^๐‘ฆ = ใ€–"(k)" ใ€—^๐‘ฆ "b" ^(๐‘ฆ ร— 1/๐‘ฆ) = "k" ^๐‘ฆ b= "k" ^๐‘ฆ Similarly, "c" ^(1/๐‘ง) = k Taking power z both sides ("c" ^(1/๐‘ง) )^๐‘ง = ใ€–"(k)" ใ€—^๐‘ง "c" ^(๐‘ง ร— 1/๐‘ง) = "k" ^๐‘ง c = "k" ^๐‘ง Thus, a = "k" ^๐‘ฅ , b = "k" ^๐‘ฆ & c = "k" ^๐‘ง It is given that a, b & c are in GP So, ratio will be the same ๐‘/๐‘Ž = ๐‘/๐‘ b2 = ac putting value of a, b & c from (1) ("k" ^๐‘ฆ )^2 = "k" ^๐‘ฅ "k" ^๐‘ง "k" ^2๐‘ฆ = "k" ^(๐‘ฅ+๐‘ง) Comparing powers 2y = x + z We need to show x, y & z are in AP i.e. we need to show that their common difference is same i.e. we need to show y โ€“ x = z โ€“ y y + y = z + x 2y = z + x And we have proved in (2) that 2y = z + x Hence, x, y & z are in A.P. Hence proved

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