Let a, b, c d be real number such that $| a - b | = 2, | b - c | = 3, | c - d | = 4.$ Then the sum of all possible value of $| a - d |$ is

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Solution

The correct option is A 9
$| a - b | + | b - c | + | c - d |$
$= a - b + b - c + c - d$
$= 2 + 3 + 4$
$= 9$

Suggest Corrections

Similar questions

a, b, c, d

a + b + c + d = \frac{4}{3}

\sqrt{\frac{1}{b + c + d}} + \sqrt{\frac{1}{c + d + a}} + \sqrt{\frac{1}{d + a + b}} + \sqrt{\frac{1}{a + b + c}}

a + b < c + d

b + c < d + e

c + d < e + a

d + e < a + b

\frac{a}{b} = \frac{2}{3}, \frac{b}{c} = \frac{1}{3}, \frac{c}{d} = \frac{1}{4}, e = \frac{a c}{b^{2} + c^{2}}

- 5

5

| a | = \sqrt{4 - \sqrt{5 - a}}, | b | = \sqrt{4 + \sqrt{5 - b}}, | c | = \sqrt{4 - \sqrt{5 + c}}, | d | = \sqrt{4 + \sqrt{5 + d}}

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