If sum to infinity of series $3 - 5 r + 7 r^{2} - 9 r 3 + .$ is $14 / 9$ , find $r$ .

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Solution

$S = 3 - 5 r + 7 r^{2} - 9 r^{3} + . . = \frac{14}{9}$
$r s = 3 r + 5 r^{2} + 7 r^{2} + . .$
$s + r s = 3 - 2 r + 2 r^{2} - 2 r^{3} + . . .$
$s (1 + r) = 3 + \frac{(- 2 r)}{1 - r}$
$\frac{14}{9} . (1 + r) = 3 + (\frac{- 2 r}{1 - r})$
$14 + 14 r = 9 (3 (1 - r)) - 2 r \times 9$
$14 + 14 r = 27 - 27 r - 18 r$
$(14 r + 14) (1 - r) = 27 - 45 r$
$14 r - 14 r^{2} + 14 - 14 r = 27 - 45 r$
$- 14 r^{2} + 45 r - 13 = 0$
$14 r^{2} - 45 r + 13 = 0$
value of roots of this eq $^{n}$

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