Question

An infinite geometric progression has a positive common ratio and third term equal to 8. The smallest possible value of its sum is

• A. 54
• B. 56
• C. 64
• D. 72

Answer 1

Answer: (A)

$r^2=8$
$\therefore a=\frac{8}{r^2};\text{ sum }=\frac{\frac{8}{r^2}}{1-r}=\frac{8}{r^2(1-r)}$
$f(r)=r^2-r^3$
$f^{\prime }(r)=2r-3r^2=r(2-3r)$
$\therefore r=\frac{2}{3}$
$f^{\prime \prime }(r)=2-6r<0\Rightarrow \text{ maxima }$
$\therefore \text{ minimum sum }=\frac{8}{\frac{4}{9}\left(1-\frac{2}{3}\right)}=\frac{8\times 9\times 3}{4\times 1}=54$