Answer:
- [tex]\dfrac{a+b+c}{a} =7[/tex]
Step-by-step explanation:
Given :-
- [tex]\dfrac{a+b}{3} =\dfrac{b+c}{6} =\dfrac{c+a}{5}[/tex]
And we need to prove that ,
- [tex]\dfrac{a+b+c}{a} =7[/tex]
So let us assume that ,
[tex]\implies\dfrac{a+b}{3} =\dfrac{b+c}{6} =\dfrac{c+a}{5}=k[/tex]
Where k is a constant . Now equate each of the three terms separately to k . Therefore we have ,
[tex]\dfrac{a+b}{3}=k[/tex]
[tex]\implies a + b = 3k \quad \dots (i) [/tex]
Similarly we can say that ,
[tex]\implies c + b = 6k \quad \dots (ii)[/tex]
[tex]\implies a + c = 5k\quad \dots (iii) [/tex]
Subtracting (i) and (ii) :-
[tex]\implies a - c = -3k \quad \dots (iv) [/tex]
Adding (iv) and (iii) :-
[tex]\implies 2c = 4k [/tex]
[tex]\implies \boxed{c = 4k }[/tex]
Put this is (ii) :-
[tex]\implies b +4k = 6k [/tex]
[tex]\implies \boxed{b= 2k }[/tex]
Similarly we will get ,
[tex]\implies \boxed{a = k }[/tex]
Proving the given equation :-
[tex]\implies \dfrac{a+b+c}{a} \\\\\implies \dfrac{k+2k+4k}{k} \\\\\implies \dfrac{7k}{k}=\boxed{\red{ 7}}[/tex]
Hence proved !
