if a , b , c are in continued proportion then show that (a+b)²/ab = (b+c)²/bc
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Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let)
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c)
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1)
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2]
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2] =c2[k4+2k2+1−k2]
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2] =c2[k4+2k2+1−k2] =c2[k4+k2+1]
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2] =c2[k4+2k2+1−k2] =c2[k4+k2+1]R.H.S. =c2k4+c
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2] =c2[k4+2k2+1−k2] =c2[k4+k2+1]R.H.S. =c2k4+cStep-by-step explanation:
Answer:To Prove : (a+b+c)(a−b+c)=a2+b2+c2Proof : a,b,c are in continued proportion.∴ ba=cb=k (let) b=ck a=bk=(ck)k =ck2L.H.S. =(ck2+ck+c)(ck2−ck+c) =c2(k2+k+1)(k2−k+1) =c2[(k2+1)2−(k)2] =c2[k4+2k2+1−k2] =c2[k4+k2+1]R.H.S. =c2k4+cStep-by-step explanation:please give me a point please make me brainless I hope I can help you good day
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