Question

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

Answer 1

a/b=b/c=k (say)
a=bk;   b=ck;   a=ck^2
L.H.S                                                          
(ck2+ck+c)(ck2-ck-c)
c^2(k2+k+1)(k^2-k+1)
c2(k4-k3+k2+k3-k2+k+k2-k+1)
c2(k4+k2+1)
RHS
c2k4+c2k2+c2
c2(k4+k2+1)=RHS

Answer 2

$\underline\mathfrak\orange{Answer:–}$

Given:

a, b and c are in continued proportion.

.

To Prove:

(a + b + c)(a - b + c) = a² + b² + c²

.

Proof:

$\begin{gathered}\text{As a, b and c are in continued proportion,} \\\\:\longrightarrow \sf{a : b : : b : c}\\\\:\implies \sf{a : b = b : c} \\\\:\implies \sf{\dfrac{a}{b} = \dfrac{b}{c}} \\\\\text{On cross multiplying,}\\\\:\implies \sf{a*c = b*b} \\\\:\implies \sf{ac = b^2----------(1)}\end{gathered}$

.

$\begin{gathered}\\\\\text{Now, we take and solve LHS:-}\\\\:\implies \sf{(a + b + c)(a - b + c)} \\\\:\implies \sf{(a^2 - ab + ac) + (ab - b^2 + bc) + (ac - bc + c^2)} \\\\:\implies \sf{a^2-{ab\!\!\!\!/}+ ac +{ab\!\!\!\!/} - b^2 + {bc\!\!\!\!/} + ac - {bc\!\!\!\!/}+ c^2} \\\\:\implies \sf{a^2 + 2ac - b^2 + c^2} \\\\\text{Using value of 'ac' from (1)} \\\\:\implies \sf{a^2 + 2(b^2) - b^2 + c^2} \\\\:\implies \sf{a^2 + 2b^2 - b^2 + c^2} \\\\:\implies \bf{a^2 + b^2 +c^2} \\\\\text{LHS = RHS} \\\\\text{Hence proved}\end{gathered}$