If (a + b), (b – c) and (c + d) are in AP, then which of the following is in AP?

\frac{a + b}{2 a}, \frac{b - c}{2 a}, \frac{c + d}{a} (a \neq 0)

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\frac{2 (a + b)}{3}, \frac{2 (b - c)}{3}, \frac{2 c + 2 d}{3}

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(a - b), (b + c), (c + d)

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\frac{a + b}{a - b}, \frac{b - c}{b + c}, \frac{c + d}{c - d}

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Solution

The correct option is B $\frac{2 (a + b)}{3}, \frac{2 (b - c)}{3}, \frac{2 c + 2 d}{3}$
Given that $(a + b), (b - c)$ and $(c + d)$ are in A.P
∴ 2(b – c) = a + b + c + d …..(i)
Now, consider:
$\frac{c + d}{a} + \frac{a + b}{2 a} = \frac{2 c + 2 d + a + b}{2 a}$
$= \frac{a + b + c + d + c + d}{2 a}$
$= \frac{2 (b - c) + c + d}{2 a}$
$= \frac{2 b - 2 c + c + d}{2 a}$
$= \frac{2 b + d - c}{2 a}$
$= \frac{2 (b - c)}{2 a}$
Thus, $\frac{a + b}{2 a}, \frac{b - c}{2 a}, \frac{c + d}{a} (a \neq 0)$ are not in A.P

Consider
$\frac{2 c + 2 d}{3} + \frac{2 (a + b)}{3} = \frac{2 c + 2 d + 2 a + 2 b}{3}$
$= \frac{2 (a + b + c + d)}{3}$
$= \frac{2 [2 (b - c)]}{3}$
$= \frac{4 (b - c)}{3} = 2 \times \frac{2 (b - c)}{3}$
Thus, $\frac{2 (a + b)}{3}, \frac{2 (b - c)}{3}, \frac{2 c + 2 d}{3}$ are in A.P
Consider,
$(c + d) + (a - b) = a - b + c + d$
$\neq 2 (b + c)$
$\frac{a + b}{a - b} + \frac{c + d}{c - d} = \frac{(a + b) (c - d) + (a - b) (c + d)}{(a - b) (c - d)}$
$= \frac{a c - a d + b c - b d + a c + a d - b c - b d}{a c - a d - b c + b d}$
$= \frac{2 a c - 2 b d}{a c - a d - b c + b d}$
$= \frac{2 (a c - b d)}{a c - a d - b c + b d}$
$\neq 2 (\frac{b - c}{b + c})$
Thus, $\frac{a + b}{a - b}, \frac{b - c}{b + c}, \frac{c + d}{c - d}$ are not in A.P

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