If abcd are in continued proportion than show √a5+b2c2+a3c2/b4c+d4+b

1.	If abcd are in continued proportion than show √a5+b2c2+a3c2/b4c+d4+b2cd2=a/d
Answer» Given: $a$ , $b$ , $c$ and $d$ are in continued proportion To prove: $\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}=\dfrac{a}{d}$ Step-by-step EXPLANATION: Since $a$ , $b$ , $c$ and $d$ are in continued proportion, $\quad\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k$ (say), $k\neq 0$ Then, we have $\quad a=bk,b=ck,c=dk$ $\Rightarrow a=dk^{3},b=dk^{2},c=dk$ Now, Left Hand Side $=\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}$ $=\sqrt{\dfrac{(dk^{3})^{5}+(dk^{2})^{2}(dk)^{2}+(dk^{3})^{3}(dk)^{2}}{(dk^{2})^{4}(dk)+d^{4}+(dk^{2})^{2}(dk)d^{2}}}$ $=\sqrt{\dfrac{d^{5}k^{15}+d^{2}k^{4}.d^{2}k^{2}+d^{3}k^{9}.d^{2}k^{2}}{d^{2}k^{<klux>8</klux>}.dk+d^{4}+d^{2}k^{4}.dk.d^{2}}}$ $=\sqrt{\dfrac{d^{5}k^{15}+d^{4}k^{6}+d^{5}k^{11}}{d^{5}k^{9}+d^{4}+d^{5}k^{5}}}$ $=\sqrt{\dfrac{d^{4}k^{6}(dk^{9}+1+dk^{5})}{d^{4}(dk^{9}+1+dk^{5})}}$ $=\sqrt{k^{6}}$ $=k^{3}$ and Right Hand Side $=\dfrac{a}{d}$ $=\dfrac{dk^{3}}{d}$ $=k^{3}$ So, Right Hand Side = Left Hand Side Conclusion: $\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}=\dfrac{a}{d}$ Proved. Read more on Brainly.in Sachin, Simon and Dhanush scored 152 runs in a T -20 match. Their scores are in a continued proportion. Simon and Dhanush together scored 8 runs more than Sachin. Find their individual scores. brainly.in/question/2695989 The ratio of three numbers is 4 : 3 : 7 and SUM of their square is 666. What is the value of the largest of the three numbers? brainly.in/question/20656096