Proof problem on ratios

If ab=bc=cd\dfrac{a}{b}=\dfrac {b}{c}=\dfrac{c}{d},prove that:ad=a5+b2c2+a3c2b4c+d4+b2cd2.\dfrac{a}{d}=\sqrt{\dfrac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}.

#Algebra

Note by Rohit Udaiwal
5 years, 5 months ago

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Comments

Let ab=bc=cd=k∴a=kb,b=kc,c=kdakd=kbc⇒ad=k2⋅bc=k2⋅k=k3Now, =a5+b2c2+a3c2b4c+d4+b2cd2Replacing a=kb,b=kc and c=kd,=(kb)5+(kc)2(kd)2+(kb)3(kd)2(kc)4(kd)+d4+(kc)2(kd)d2Replacing b=kc and c=kd again,=(k(kc))5+(k(kd))2(kd)2+(k(kc))3(kd)2(k(kd))4(kd)+d4+(k(kd))2(kd)d2Replacing c=kd again,=(k(k(kd)))5+(k(kd))2(kd)2+(k(k(kd)))3(kd)2(k(kd))4(kd)+d4+(k(kd))2(kd)d2=(k3d)5+(k2d)2(kd)2+(k3d)3(kd)2(k2d)4(kd)+d4+(k2d)2(kd)d2=k15d5+k6d4+k11d5k9d5+d4+k5d5=k6(k9d5+d4+k5d5)k9d5+d4+k5d5=k6=k3\text{Let } \frac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k \\ \therefore a=kb,b=kc,c=kd \\ \dfrac{a}{kd}=\dfrac{kb}{c}\Rightarrow \dfrac{a}{d}=k^2 \cdot \frac{b}{c}=k^2 \cdot k =\boxed{k^3} \\ \text{Now, } \\ =\sqrt{\dfrac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}\\ \text{Replacing a=kb,b=kc and c=kd,} \\ =\sqrt{\dfrac{(kb)^5+(kc)^2 (kd)^2+(kb)^3 (kd)^2}{(kc)^4 (kd)+d^4+(kc)^2 (kd) d^2}} \\ \text{Replacing b=kc and c=kd again,} \\ =\sqrt{\dfrac{(k(kc))^5+(k(kd))^2 (kd)^2+(k(kc))^3 (kd)^2}{(k(kd))^4 (kd)+d^4+(k(kd))^2 (kd) d^2}} \\ \text{Replacing c=kd again,} \\ =\sqrt{\dfrac{(k(k(kd)))^5+(k(kd))^2 (kd)^2+(k(k(kd)))^3 (kd)^2}{(k(kd))^4 (kd)+d^4+(k(kd))^2 (kd) d^2}} \\ =\sqrt{\dfrac{(k^3 d)^5+(k^2 d)^2 (kd)^2+(k^3 d)^3 (kd)^2}{(k^2 d)^4 (kd)+d^4+(k^2 d)^2 (kd) d^2}} \\ =\sqrt{\dfrac{k^{15}d^5+k^6 d^4+k^{11}d^5}{k^9 d^5+d^4+k^5 d^5}} \\ =\sqrt{\dfrac{k^6(k^9 d^5+d^4+k^5 d^5)}{k^9 d^5+d^4+k^5 d^5}} \\ =\sqrt{k^6}=\boxed{k^3}

Akshat Sharda - 5 years, 5 months ago

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Nice solution bro!😊

Akshay Yadav - 5 years, 5 months ago

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Thanks brother !!

Akshat Sharda - 5 years, 5 months ago
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