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


Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.