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(B) Let the first term be $a = 1$ and the common ratio be $r$.The terms of the geometric progression are $T_1 = 1$,$T_2 = r$,and $T_3 = r^2$.We are gi

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$-\frac{2}{5}$

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(B) Let the first term be $a = 1$ and the common ratio be $r$.The terms of the geometric progression are $T_1 = 1$,$T_2 = r$,and $T_3 = r^2$.We are given the expression $f(r) = 4T_2 + 5T_3 = 4r + 5r^2$.To find the minimum value,we differentiate $f(r)$ with respect to $r$ and set it to $0$:$f'(r) = \frac{d}{dr}(4r + 5r^2) = 4 + 10r$.Setting $f'(r) = 0$ gives $4 + 10r = 0$,which implies $10r = -4$.Thus,$r = -\frac{4}{10} = -\frac{2}{5}$.Since $f''(r) = 10 > 0$,the function has a minimum at $r = -\frac{2}{5}$.

In a geometric progression,the first term is $1$. If $4T_2 + 5T_3$ is minimum,then what is the common ratio?

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