Question

If p, q, r, s are in G.P. then show that
( q-r)2 + ( r-p ) 2
+ (s-q)2
= (p-s)2

Answer 1

Step-by-step explanation:

(q-r)2+(r-p)2+(s-q)2= 2q-2r+2r-2p+2s-2q

= 2s-2p

=(s-p)2

Answer 2

Answer:

The relation $(q-r)^{2}+(r-p)^{2}+(s-q)^{2}=(p-s)^{2}$ is proved in the below answer.

Step-by-step explanation:

Given that,

p,q,r,s are in G.P and we are required to prove that

$(q-r)^{2}+(r-p)^{2}+(s-q)^{2}=(p-s)^{2}$

Let the common difference of the given geometric progression be k.

Therefore,the other terms become

$q=pk,r=pk^{2},s=pk^{3}$

Now solving for given expression,

$L.H.S=(q-r)^{2}+(r-p)^{2}+(s-q)^{2}\\=(pk-pk^{2})^{2}+(pk^{2}-p)^{2}+(pk^{3}-pk^{2})^{2}\\=p^{2}k^{2}(1-k)^{2}+p^{2}(k^{2}-1)^{2}+p^{2}k^{4}(k-1)^{2}\\=p^{2}k^{2}(1-k)^{2}+p^{2}[(1+k)(1-k)]^{2}+p^{2}k^{4}(k-1)^{2}\\=p^{2}(1-k)^{2}[k^{2}+(1+k)^{2}+k^{4}]\\=p^{2}(1-k)^{2}[k^{4}+2k^{2}+2k+1]=p^{2}(1-k^{3})^{2}$

Similarly,

$R.H.S=(p-s)^{2}=(p-pk^{3})^{2}=p^{2}(1-k^{3})^{2}=L.H.S\\\therefore L.H.S=R.H.S$

Therefore,

The relation $(q-r)^{2}+(r-p)^{2}+(s-q)^{2}=(p-s)^{2}$ is proved.

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