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Q.
An infinite geometric progression has a positive common ratio and third term equal to 8. The smallest
possible value of its sum is
Sequences and Series
Solution:
$r ^2=8 $
$\therefore a =\frac{8}{ r ^2} ; \text { sum }=\frac{\frac{8}{r^2}}{1-r}=\frac{8}{ r ^2(1-r)} $
$f ( r )= r ^2- r ^3$
$f ^{\prime}( r )=2 r -3 r ^2= r (2-3 r ) $
$\therefore r =\frac{2}{3}$
$f ^{\prime \prime}( r )=2-6 r <0 \Rightarrow \text { maxima } $
$\therefore \text { minimum sum }=\frac{8}{\frac{4}{9}\left(1-\frac{2}{3}\right)}=\frac{8 \times 9 \times 3}{4 \times 1}=54$