If the first three terms of the sequence frac | vedclass

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

Question

(C) If $\frac{1}{16}, a, b$ are in $G.P.$,then $a^2 = \frac{b}{16}$,which implies $b = 16a^2$ ..... $(i)$If $a, b, \frac{1}{6}$ are in $H.P.$,then $b

Vedclass · Accepted Answer

Both $(a)$ and $(b)$ are true

Vedclass · Answer

(C) If $\frac{1}{16}, a, b$ are in $G.P.$,then $a^2 = \frac{b}{16}$,which implies $b = 16a^2$ ..... $(i)$If $a, b, \frac{1}{6}$ are in $H.P.$,then $b = \frac{2 \cdot a \cdot \frac{1}{6}}{a + \frac{1}{6}} = \frac{2a}{6a + 1}$ ..... $(ii)$Equating $(i)$ and $(ii)$,we get $16a^2 = \frac{2a}{6a + 1}$.For $a 
eq 0$,$8a = \frac{1}{6a + 1}$,which gives $48a^2 + 8a - 1 = 0$.Factoring the quadratic equation: $(4a + 1)(12a - 1) = 0$.Thus,$a = -\frac{1}{4}$ or $a = \frac{1}{12}$.If $a = -\frac{1}{4}$,then $b = 16(-\frac{1}{4})^2 = 1$.If $a = \frac{1}{12}$,then $b = 16(\frac{1}{12})^2 = 16 \cdot \frac{1}{144} = \frac{1}{9}$.Therefore,both pairs are valid solutions.

If the first three terms of the sequence $\frac{1}{16}, a, b, \frac{1}{6}$ are in a geometric progression and the last three terms are in a harmonic progression,then the values of $a$ and $b$ are:

Similar Questions

If $a_n = \sqrt{7+\sqrt{7+\sqrt{7+\ldots}}}$ ($n$ times),then which one of the following is true?

What is the sum of the first $40$ terms of the series $1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 + 9 + \dots$?

Let $a_1, a_2, a_3, \ldots$ be an $A$.$P$. If $a_7 = 3$,the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero,then $n! - 4 a_{n(n+2)}$ is equal to :

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

Define a sequence $\{a_n\}_{n \geq 0}$ by $a_n = \sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1$,with $a_0 = \cos \theta \neq \pm 1$. Then,$\lim_{n \rightarrow \infty} 4^n(1-a_n)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module