Question

Consider a geometric series: $1,b_2,b_3,...$. The minimum value of $4b_2+5b_3$, is

• A. $-\frac{2}{5}$
• B. $-\frac{4}{5}$
• C. $-\frac{1}{5}$
• D. $-\frac{8}{5}$

Answer 1

Answer: (B)

$-\frac{4}{5}$

Let the common ratio of the G.P. is r.
$b_1=1$
$b_2=r$
$b_3=r^2$
$\therefore 4b_2+5b_3$
$=4r+5r^2=5(r^2+\frac{4}{5}r)=5(r^2+2.\frac{2}{5}r+{\left(\frac{2}{5}\right)}^2-{\left(\frac{2}{5}\right)}^2)=5[{(r+\frac{2}{5})}^2-\frac{4}{25}]=5{(r+\frac{2}{5})}^2-\frac{4}{5}$
The first term of the expression is a square term, which has minimum value as zero.
Hence, the minimum value of the expression is $-\frac{4}{5}$