If $a, b, c, d, e$ are in continued proportion, prove that
$(a b + b c + c d + d e)^{2} = (a^{2} + b^{2} + c^{2} + d^{2}) (b^{2} + c^{2} + d^{2} + e^{2})$ .

Open in App

Solution

$a, b, c, d, e$ are in continued proportion,

$\Rightarrow \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e} = k \Rightarrow a = b k, b = c k, c = d k, d = e k$

We need to prove ${(a b + b c + c d + d e)}^{2} = (a^{2} + b^{2} + c^{2} + d^{2}) (b^{2} + c^{2} + d^{2} + e^{2})$ ....(1)

Substituting the values of $a, b, c, d$ in (1)

$\Rightarrow {(b^{2} k + c^{2} k + d^{2} k + e^{2} k)}^{2} = ({(b k)}^{2} + {(c k)}^{2} + {(d k)}^{2} + {(e k)}^{2}) (b^{2} + c^{2} + d^{2} + e^{2}) \Rightarrow k^{2} {(b^{2} + c^{2} + d^{2} + e^{2})}^{2} = k^{2} (b^{2} + c^{2} + d^{2} + e^{2}) (b^{2} + c^{2} + d^{2} + e^{2}) \Rightarrow {(b^{2} + c^{2} + d^{2} + e^{2})}^{2} = {(b^{2} + c^{2} + d^{2} + e^{2})}^{2}$
As LHS $=$ RHS

Hence, proved.

Suggest Corrections

Similar questions

a, b, c, d

(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a b + b c + c d)^{2}

a, b, c

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{2} + b^{2} + c^{2}}{{(a + b + c)}^{2}} = \frac{a - b + c}{a + b + c}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}}

Join BYJU'S Learning Program