Question

if a b c and d are in proportion prove that abcd(1/a²+1/b²+1/c²+1/d²)=a²+b²+c²+d²​

Answer 1

GIVEN :–

• a,b,c and d are in proportion.

TO PROVE :–

• abcd(1/a²+1/b²+1/c²+1/d²)=a²+b²+c²+d²

SOLUTION :–

• According to given condition –

$\\\bf \implies\dfrac{a}{b} = \dfrac{c}{d} = k(const)$

$\\\bf \implies \: a= bk \: \: \& \: \: c = dk$

• Now let's take L.H.S. –

$\\\bf \: \: = \: \: abcd \left( \dfrac{1}{a^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{c^{2}}+ \dfrac{1}{d^{2}} \right)$

• Now put the values –

$\\\bf \: \: = \: \: (bk)b(dk)d \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\bf \: \: = \: \: b^{2} {d}^{2} k^{2} \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\bf \: \: = \: \: b^{2} {d}^{2} k^{2} \left[\dfrac{ {d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2} }{b^{2}k^{2}d^{2}} \right]$

$\\\bf \: \: = \: \:{d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2}$

$\\\bf \: \: = \: \:{d}^{2} +{(dk)}^{2} + {b}^{2} + {(bk)}^{2}$

$\\\bf \: \: = \: \:{d}^{2} +c^{2} + {b}^{2} +a^{2}$

$\\\bf \: \: = \: \:R.H.S.$

$\\\bf \longrightarrow \: \: Hence \: \: proved$

Answer 2

Answer:

Here is your answer mate please refer to the attachment.