AP and GP mix questions

Chapter 9 Class 11 Sequences and Series (Term 1)
Concept wise

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Example 22 If a, b, c are in G.P. and "a" ^(1/π₯) = "b" ^(1/π¦) = "c" ^(1/π§) , prove that x, y, z are in A.P. Given that "a" ^(1/π₯) = "b" ^(1/π¦) = "c" ^(1/π§) Let "a" ^(1/π₯) = "b" ^(1/π¦) = "c" ^(1/π§) = k Now, "a" ^(1/π₯) = k Taking power x both sides ("a" ^(1/π₯) )^π₯ = γ"(k)" γ^π₯ "a" ^(π₯ Γ 1/π₯) = "k" ^π₯ a = "k" ^π₯ Also, "b" ^(1/π¦) = k Taking power y both sides ("b" ^(1/π¦) )^π¦ = γ"(k)" γ^π¦ "b" ^(π¦ Γ 1/π¦) = "k" ^π¦ b= "k" ^π¦ Similarly, "c" ^(1/π§) = k Taking power z both sides ("c" ^(1/π§) )^π§ = γ"(k)" γ^π§ "c" ^(π§ Γ 1/π§) = "k" ^π§ c = "k" ^π§ Thus, a = "k" ^π₯ , b = "k" ^π¦ & c = "k" ^π§ It is given that a, b & c are in GP So, ratio will be the same π/π = π/π b2 = ac putting value of a, b & c from (1) ("k" ^π¦ )^2 = "k" ^π₯ "k" ^π§ "k" ^2π¦ = "k" ^(π₯+π§) 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 β x = z β 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