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In-the-binomial-expasion-of-a-b-5-the-sum-of-2-nd-and-3-rd-term-is-zero-then-a-b-is-




Question Number 22612 by Tinkutara last updated on 21/Oct/17
In the binomial expasion of (a − b)^5 ,  the sum of 2^(nd)  and 3^(rd)  term is zero,  then (a/b) is
$${In}\:{the}\:{binomial}\:{expasion}\:{of}\:\left({a}\:−\:{b}\right)^{\mathrm{5}} , \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term}\:{is}\:{zero}, \\ $$$${then}\:\frac{{a}}{{b}}\:{is} \\ $$
Answered by ajfour last updated on 21/Oct/17
⇒ −5a^4 b+10a^3 b^2 =0  ⇒ (a/b)=2 .
$$\Rightarrow\:−\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\frac{{a}}{{b}}=\mathrm{2}\:. \\ $$
Commented by Tinkutara last updated on 21/Oct/17
But in book answer given is (1/2). Is this  wrong?
$${But}\:{in}\:{book}\:{answer}\:{given}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}.\:{Is}\:{this} \\ $$$${wrong}? \\ $$
Answered by ajfour last updated on 21/Oct/17
(a−b)^5 =−(b−a)^5   sum of 2^(nd)  and 3^(rd)  term       =5b^4 a−10b^3 a^2 =0  ⇒    (a/b)=(1/2) .
$$\left({a}−{b}\right)^{\mathrm{5}} =−\left({b}−{a}\right)^{\mathrm{5}} \\ $$$${sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term} \\ $$$$\:\:\:\:\:=\mathrm{5}{b}^{\mathrm{4}} {a}−\mathrm{10}{b}^{\mathrm{3}} {a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\frac{{a}}{{b}}=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

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