If x-a-ar-ar2-y-b-br-br2- andz-c-cr2-cr4- r-1- c-0then xyz-

X=a+ar+ar2+....∞, nbsp;⇒x=a1 1r=arr 1y=b br+br2 ....∞⇒y=b1–1r=brr+1z=c+cr2+cr4+....∞, nbsp;⇒z=c1 1r2=cr2r2 1⇒zc=r2r2 1 nbsp;..... nbsp;1xy=arr 1×brr+1=a ...

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Byju's Answer

Standard XI

Mathematics

Sum of Infinite Terms

If x=a+ar+ar2...

Question

If $x = a + \frac{a}{r} + \frac{a}{r^{2}} + . . . . \infty, y = b - \frac{b}{r} + \frac{b}{r^{2}} - . . . . \infty, a n d z = c + \frac{c}{r^{2}} + \frac{c}{r^{4}} + . . . . \infty, (r | > 1, c \neq 0 t h e n \frac{x y}{z} =$
.

\frac{a b}{c}

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\frac{a c}{b}

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abc

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\frac{c}{a b}

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Solution

The correct option is A $\frac{a b}{c}$
$x = a + \frac{a}{r} + \frac{a}{r^{2}} + . . . . \infty, \Rightarrow x = \frac{a}{(1 - \frac{1}{r})} = \frac{a r}{r - 1} y = b - \frac{b}{r} + \frac{b}{r^{2}} - . . . . \infty \Rightarrow y = \frac{b}{(1 - (- \frac{1}{r}))} = \frac{b r}{r + 1} z = c + \frac{c}{r^{2}} + \frac{c}{r^{4}} + . . . . \infty, \Rightarrow z = \frac{c}{(1 - \frac{1}{r^{2}})} = \frac{c r^{2}}{r^{2} - 1} \Rightarrow \frac{z}{c} = \frac{r^{2}}{r^{2} - 1} . . . . . (1] x y = \frac{a r}{r - 1} \times \frac{b r}{r + 1} = a b \frac{r^{2}}{r^{2} - 1} \Rightarrow x y = a b \times \frac{z}{c} \Rightarrow \frac{x y}{z} = \frac{a b}{c}$

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If x=a+ar+ar2+....∞,y=b−br+br2−....∞, andz=c+cr2+cr4+....∞, (r|>1, c≠0then xyz= .

The correct option is A abcx=a+ar+ar2+....∞, ⇒x=a(1−1r)=arr−1y=b−br+br2−....∞⇒y=b(1−(−1r))=brr+1z=c+cr2+cr4+....∞, ⇒z=c(1−1r2)=cr2r2−1⇒zc=r2r2−1 ..... (1]xy=arr−1×brr+1=abr2r2−1⇒xy=ab×zc⇒xyz=abc

If $x = a + \frac{a}{r} + \frac{a}{r^{2}} + . . . . \infty, y = b - \frac{b}{r} + \frac{b}{r^{2}} - . . . . \infty, a n d z = c + \frac{c}{r^{2}} + \frac{c}{r^{4}} + . . . . \infty, (r | > 1, c \neq 0 t h e n \frac{x y}{z} =$
.