Maths · Arithmetic and Geometric progressions
f are in A.P then , b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\f
f \( a, b, c \) are in A.P then \( a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right) \) are in
- A. \( A . P . \)
- B. G.P.
- C. \( H . P \)
- D. A.G.P
Step-by-step solution
Given a, b, c are in A.P., so 2b = a + c. The expressions simplify to T1 = a/b + a/c, T2 = b/c + b/a, T3 = c/a + c/b. Then T1 + T3 = (a+c)/b + (a/c + c/a) = 2 + (a/c + c/a) and 2T2 = 2(b/c + b/a) = (a+c)/c + (a+c)/a = a/c + 1 + 1 + c/a = a/c + c/a + 2. Hence T1 + T3 = 2T2, so they are in A.P.
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