Question

if a,b,c are in continued proportion,prove that (a+b+c) (a-b+c) = a²+b²+c²​

Answer 1

Step-by-step explanation:

given, that a,b,c are in continued proportion,

=> a:b:c

if, k be the proportionality constant,

then, $\frac{a}{b}$ = $\frac{b}{c}$ = k

=> a = bk  and b = ck

so, a = (ck)*k

     a = ck²

we need to prove that, (a+b+c) (a-b+c) = a²+b²+c²​

we take the LHS = (a+b+c) (a-b+c)

= (ck²+ ck+ c)(ck²- ck+ c)

= c (k²+ k+ 1)*c(k²- k+ 1)

= c² (k²+ 1+ k)(k²+ 1 - k)

= c² [ (k²+ 1)² - k² ]

= c² [ (k²+ 1)² - k² ]

= c² [ k⁴ +2k²+ 1 - k² ]

= c² [ k⁴ +k²+ 1 ]

= (ck²)² +(ck)²+ c²

= a²+b²+c²

= RHS

=> LHS = RHS

Hence proved.