Web30 mrt. 2024 · 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. Web5 apr. 2024 · Hint: For a relation between a, b, c using the formula of arithmetic mean given as 2 b = a + c. For the terms b, c, d in GP use the formula of geometric mean given as d …
If a, b, c, d are in GP, prove that `(b-c)^(2)+(c-a)^(2)+(d-b
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If a, b, c, d are in GP, how can you show that a + d > b + c?
Web22 feb. 2024 · The given polynomial can be rewritten as factors of the common ratio of the geometric progression.Response:If a, b, c, d are in GP, (a² + b² + c²) × (b² + c² + … Web7 sep. 2024 · Here, it is given that a, b, c are in G.P ⇒ c = ar × r ⇒ c = ar2 So, Ans. For a, b, c to be in G.P. the value of (a - b)/ (b - c) is equal to a/b or b/c. ← Prev Question Next … Web9 dec. 2015 · 1 Just write everything in terms of a and b : c = 2b − a d = (2b − a)2 b 1 e = 2 d − 1 c Therefore, substituting c and d from above: e = (2b − a)2 a Now it is easy to see that the sequence a, c, e is a G.P: (a, c, e) = (a, 2b − a, (2b − a)2 a) therefore √ae = c. Share Cite Follow edited Dec 9, 2015 at 8:54 answered Dec 9, 2015 at 8:40 Seven otc healthnet