Question

Question 6
(a)
If a, b, c are in continued proportion, prove that
(a + b + c) (a - b + c) = a2 + b2 + c2.​

Answer 1

ANSWER:

Given:

• a, b and c are in continued proportion.

To Prove:

• (a + b + c)(a - b + c) = a² + b² + c²

Proof:

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

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