JEE Main 2017 — Mathematics Algebra

$$225{a}^{2}+9{b}^{2}+25{c}^{2}-75ac-45ab-15bc=0$$225a2+9b2+25c2−75ac−45ab−15bc=0

$$\Rightarrow 2({(15a)}^{2}+{(3b)}^{2}+{(5c)}^{2}-(15a)(5c)-(15a)(3b)-(3b)(5c)=0 )$$⇒2((15a)2+(3b)2+(5c)2−(15a)(5c)−(15a)(3b)−(3b)(5c)=0)

$$\Rightarrow 2{(15a)}^{2}+2{(3b)}^{2}+2{(5c)}^{2}-2(15a)(5c)-2(15a)(3b)-2(3b)(5c)=0$$⇒2(15a)2+2(3b)2+2(5c)2−2(15a)(5c)−2(15a)(3b)−2(3b)(5c)=0

$$\Rightarrow {(15a-3b)}^{2}+{(3b-5c)}^{2}+{(5c-15a)}^{2}=0$$⇒(15a−3b)2+(3b−5c)2+(5c−15a)2=0

$$\Rightarrow 15a-3b=3b-5c=5c-15a=0$$⇒15a−3b=3b−5c=5c−15a=0

$$\Rightarrow 15a=3b=5c$$⇒15a=3b=5c

$$\Rightarrow \frac{a}{1}=\frac{b}{5}=\frac{c}{3}=k$$⇒1a​=5b​=3c​=k

$$\Rightarrow a=k, b=5k, c=3k$$⇒a=k,b=5k,c=3k

$$\Rightarrow a, c, b are in A.P.$$⇒a,c,bareinA.P.

$$\Rightarrow b, c, a are in A.P.$$⇒b,c,aareinA.P.