If $a, b, c$ are in AP, $b - a, c - b$ and $a$ are in GP, then $a : b : c$ is

1 : 2 : 3

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1 : 3 : 5

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2 : 3 : 5

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1 : 2 : 4

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Solution

The correct option is A $1 : 2 : 3$
Given, $a, b, c$ are in AP and $b - a, c - b, a$ are in GP.
Therefore, $2 b = a + c$ and $(c - b)^{2} = (b - a) a$
$\Rightarrow (b - a)^{2} = (b - a) a$
$\Rightarrow b = 2 a$
$\Rightarrow c = 3 a$
Thus, $a : b : c = 1 : 2 : 3$ .

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Similar questions

If $a, b, c$ are in A.P. and $a, c - b, b - a$ are in G.P., where $a \neq b \neq c$ , then $a : b : c$ is ___.

a, b, c

a, c - b, b - a

(a \neq b \neq c),

a : b : c

a : b = \frac{1}{2} : \frac{1}{3}

c : b = \frac{1}{5} : \frac{1}{4}

a, b, c

a^{2}, b^{2}, c^{2}

a < b < c

a + b + c =

\frac{3}{2}

a

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