If a, b, c are in A.P. and a^2, b^2, c^2 are | vedclass

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

Question

(D) Since $a, b, c$ are in $A.P.$,we have $a+c = 2b$.Given $a+b+c = \frac{3}{4}$,substituting $a+c = 2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$

Vedclass · Accepted Answer

$\frac{1}{4} - \frac{1}{2\sqrt{2}}$

Vedclass · Answer

(D) Since $a, b, c$ are in $A.P.$,we have $a+c = 2b$.Given $a+b+c = \frac{3}{4}$,substituting $a+c = 2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$.Since $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2c^2$,which implies $ac = \pm b^2 = \pm \frac{1}{16}$.Since $a We have $a+c = 2b = \frac{1}{2}$ and $ac = -\frac{1}{16}$.These are roots of the quadratic equation $x^2 - (a+c)x + ac = 0$,which is $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$.Multiplying by $16$,we get $16x^2 - 8x - 1 = 0$.Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$,we get $x = \frac{8 \pm \sqrt{64 - 4(16)(-1)}}{32} = \frac{8 \pm \sqrt{128}}{32} = \frac{8 \pm 8\sqrt{2}}{32} = \frac{1}{4} \pm \frac{\sqrt{2}}{4} = \frac{1}{4} \pm \frac{1}{2\sqrt{2}}$.Since $a < b$,we choose the smaller root: $a = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $a < b < c$ and $a+b+c = \frac{3}{4}$,then the value of $a$ is

Similar Questions

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178.$ If the first term of this $A.P.$ is $10,$ then the median of the $A.P.$ is

The ratio of the sums of the first $n$ even numbers and $n$ odd numbers is

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

If $x \in (0, \frac{\pi}{4})$,then the expression $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ cannot take the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module