Check sibling questions

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

Example 22  - Chapter 9 Class 11 Sequences and Series - Part 2
Example 22  - Chapter 9 Class 11 Sequences and Series - Part 3
Example 22  - Chapter 9 Class 11 Sequences and Series - Part 4

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Example 22 If a, b, c are in G.P. and "a" ^(1/π‘₯) = "b" ^(1/𝑦) = "c" ^(1/𝑧) , prove that x, y, z are in A.P. Given that "a" ^(1/π‘₯) = "b" ^(1/𝑦) = "c" ^(1/𝑧) Let "a" ^(1/π‘₯) = "b" ^(1/𝑦) = "c" ^(1/𝑧) = 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/𝑦) = k Taking power y both sides ("b" ^(1/𝑦) )^𝑦 = γ€–"(k)" γ€—^𝑦 "b" ^(𝑦 Γ— 1/𝑦) = "k" ^𝑦 b= "k" ^𝑦 Similarly, "c" ^(1/𝑧) = k Taking power z both sides ("c" ^(1/𝑧) )^𝑧 = γ€–"(k)" γ€—^𝑧 "c" ^(𝑧 Γ— 1/𝑧) = "k" ^𝑧 c = "k" ^𝑧 Thus, a = "k" ^π‘₯ , b = "k" ^𝑦 & c = "k" ^𝑧 It is given that a, b & c are in GP So, ratio will be the same 𝑏/π‘Ž = 𝑐/𝑏 b2 = ac putting value of a, b & c from (1) ("k" ^𝑦 )^2 = "k" ^π‘₯ "k" ^𝑧 "k" ^2𝑦 = "k" ^(π‘₯+𝑧) 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 – x = z – 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.