Question

plz tel question maths question ​

Answer 1

Answer:

$hence \: \: \: \: \: \: a \: \: \: \: \: \: b \: \: \: \: c \: \: \: are \: \: in \: continued \: propotion$

Step-by-step explanation:

$\frac{ {(a + b)}^{2} }{ab} = \frac{ {(b + c)}^{2} }{bc} \\ \\ \\ \\ \\ \frac{ {a}^{2} + {b}^{2} + ab}{ab} = \frac{ {b}^{2} + {c}^{2} + bc }{bc} \\ \\ \\ \\ \frac{a}{b} + \frac{b}{c} + 1 = \frac{b}{c} + \frac{c}{b} + 1 \\ \\ \frac{a}{b} + \frac{b}{c} = \frac{b}{c} + \frac{c}{b} \\ \\ \frac{a}{b} = \frac{c}{b } \\ \\ \\ \\ or \: \: \: \: \: \frac{a}{b} = \frac{b}{c} \\ \\ \\ \\ hence \: \: \: \: \: \: a \: \: \: \: \: \: b \: \: \: \: c \: \: \: are \: \: in \: continued \: propotion.$