\(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\left(1\right)\)
\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\left(2\right)\)
Từ (1) và (2) => \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)
=> \(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}\)
Theo t/c dãy tsbn:
\(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=-\frac{30}{15}=-2\)
=> a/21 = -2 => a = -42
=> b/14 = -2 => b = -28
=> c/10 = -2 => c = -20
Vậy a + b + c =-42 - 28 - 20 = -90.