If the first term of a G.P. a_1, a_2, a_3, do | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that the first term $a_1 = 1$ and the common ratio is $r$.Thus,the terms are $a_1 = 1$,$a_2 = r$,and $a_3 = r^2$.We want to minimize the exp

Vedclass · Accepted Answer

$-\frac{2}{5}$

Vedclass · Answer

(A) Given that the first term $a_1 = 1$ and the common ratio is $r$.Thus,the terms are $a_1 = 1$,$a_2 = r$,and $a_3 = r^2$.We want to minimize the expression $f(r) = 4a_2 + 5a_3 = 4r + 5r^2$.To find the minimum,we take the derivative with respect to $r$ and set it to zero:$f'(r) = \frac{d}{dr}(4r + 5r^2) = 4 + 10r$.Setting $f'(r) = 0$ gives $4 + 10r = 0$,which implies $10r = -4$.Therefore,$r = -\frac{4}{10} = -\frac{2}{5}$.Since the second derivative $f''(r) = 10 > 0$,the function has a minimum at $r = -\frac{2}{5}$.

If the first term of a $G.P.$ $a_1, a_2, a_3, \dots$ is unity such that $4a_2 + 5a_3$ is least,then the common ratio of the $G.P.$ is

Similar Questions

If the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ are in a geometric progression,then $q^2+9 p^2+6 p q+q/p=$

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c$ and $d$ are in:

If $a, b, c$ are in geometric progression,then which of the following is true?

If $G_1$ and $G_2$ are two geometric means between two numbers and $A$ is the arithmetic mean between them,then find the value of $\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1}$.

In an increasing geometric progression of positive terms,the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module