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

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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 = (𝑐 + 𝑎)/2 Also given that b, c, d are in GP So, their common ratio is same 𝑐/𝑏 = 𝑑/𝑐 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 = (𝑎 + 𝑐)/2 & d = 2𝑐𝑒/(𝑐 + 𝑒) c2 = ((a + c)/2)(2ce/(c + e)) c2 = ((a + c)(2ce))/(2(c + e)) c2 = ((a + c)(ce))/((c + e)) 𝑐2/𝑐 = (𝑒(𝑎 + 𝑐))/(𝑐 + 𝑒) c = (𝑒(𝑎 + 𝑐))/(𝑐 + 𝑒) 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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