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ICSE Class X Math 2015 Paper

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Question : 23 of 52
Marks: +1, -0
If a,b,ca, b, c are in continued proportion, prove that
(a+b+c)(a−b+c)=a2+b2+c2  .   (a+b+c)(a-b+c)=a^2+b^2+c^2 \; \text{. } \;
Solution:  
To Prove : (a+b+c)(a−b+c)=a2+b2+(a+b+c)(a-b+c)=a^2+b^2+ c2c^2 .
Proof : a,b,ca, b, c are in continued proportion.
∴    ab=bc=k   (let)   \therefore \; \; \frac{a}{b} = \frac{b}{c} = k \; \text{ (let) } \;
b  =ckb \; = c k
a  =bk=(ck)ka \; = b k = (c k) k
  =ck2\; = c k^2
   L.H.S.     =(ck2+ck+c)(ck2−ck+c)\; \text{ L.H.S. } \; \; = (c k^2 + c k + c) (c k^2 - c k + c)
  =c2(k2+k+1)(k2−k+1)\; = c^2 (k^2 + k + 1) (k^2 - k + 1)
  =c2[(k2+1)2−(k)2]\; = c^2 [ (k^2+1)^2 - (k)^2 ]
  =c2[k4+2k2+1−k2]\; = c^2 [k^4 + 2 k^2 + 1 - k^2]
  =c2[k4+k2+1]\; = c^2 [k^4 + k^2 + 1]
   R.H.S.     =c2k4+c2k2+c2\; \text{ R.H.S. } \; \; = c^2 k^4 + c^2 k^2 + c^2
  =c2[k4+k2+1]\; = c^2 [k^4 + k^2 + 1]
L.H.S. = R.H.S.
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