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Misc 20 - If a, b, c are in AP, b, c, d are in GP and 1/c, 1/d - AP and GP mix questions

Misc 20 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 20 - Chapter 9 Class 11 Sequences and Series - Part 3 Misc 20 - Chapter 9 Class 11 Sequences and Series - Part 4


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Misc 20 If a, b, c are in A.P, ; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P. It is given that a, b, c are in AP So, their common difference is same b a = c b b + b = c + a 2b = c + a b = ( + )/2 Also given that b, c, d are in GP So, their common ratio is same / = / c2 = bd Also 1/c, 1/d, 1/e are in A.P. So, their common difference is same 1/d 1/c= 1/e 1/d 1/d + 1/d = 1/e + 1/c 2(1/d) = (c + e)/ec 2/d = (c + e)/ec d/2 = ec/(c + e) d = 2(ec/(c+e)) We need to show that a, c, e are in GP i.e. we need to show their common ratio is same c/a = e/c c2 = ae So, we need to show c2 = ae From (2), we have c2 = bd Putting value of b = ( + )/2 & d = 2 /( + ) c2 = ((a + c)/2)(2ce/(c + e)) c2 = ((a + c)(2ce))/(2(c + e)) c2 = ((a + c)(ce))/((c + e)) 2/ = ( ( + ))/( + ) c = ( ( + ))/( + ) c(c + e) = e(a + c) c2 + ec = ea + ec c2 = ea + ec ec c2 = ea + 0 c2 = ea Which is what we need to prove Hence proved Thus, a, c & e are in GP

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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.