Single Choice

If $$a, b, c$$ be in G.P. & $$log_c\, a,\, log_b\, c,log_a\,b$$ be in A.P., then show that the common difference of the A.P is

A$$\dfrac{1}{2}$$
B$$\dfrac{3}{2}$$
Correct Answer
C$$\dfrac{5}{2}$$
D$$\dfrac{1}{3}$$

Solution

Given $$a,b,c$$ are in GP and
$$\log{_c}a,\log{_b}c,\log{_a}b$$ are in AP
Let $$r$$ be the common ratio, then $$b=a,c=ar^2\quad (1)$$
and $$\log{_c}a,\log{_b}c,\log{_a}b$$ are in AP
i.e. $$\cfrac{\log a}{\log c},\cfrac{\log c}{\log b},\cfrac{\log b}{\log a}$$ are in AP
From $$(1)$$ $$\cfrac{\log a}{\log (ar^2)},\cfrac{\log (ar^2)}{\log (ar)},\cfrac{\log (ar)}{\log a}$$ are in AP
$$\cfrac{\log a}{\log a+2\log r},\cfrac{\log a+2\log r}{\log a+\log r},\cfrac{\log a+\log r}{\log a}$$ are in AP
Put $$\cfrac{\log r}{\log a}=x$$ we get
$$\cfrac{1}{1+2x},\cfrac{1+2x}{1+x},1+x$$ are in AP
$$\cfrac{2(1+2x)}{(1+x)}=(1+x)+\cfrac{1}{(1+2x)} \\ 2x^3-3x^2-3x=0 \\ x=\cfrac{1}{4}(3+\sqrt{3}) \quad (2) \\ 2d=(1+x)-\cfrac{1}{1+2x}$$
From $$(2)$$ $$2d=3 \Rightarrow d=\cfrac{3}{2}$$

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