Question

If first three terms of sequence $\dfrac{1}{{16}},a,b,\dfrac{1}{6}$ are in geometric sequence and last three terms are in harmonic sequence , then the value of \[a\] and $b$ will be
A) $a = - \dfrac{1}{4},b = 1$
B) \[a = \dfrac{1}{{12}},b = \dfrac{1}{9}\]
C) Option (A) and (B) both are true
D) None of these

Answer 1

Hint: A geometric sequence is a sequence for which the ratio of each two consecutive terms is a constant function and harmonic progression is a progression formed by taking the reciprocal of an arithmetic progression . Properties of geometric progression are , there are three terms in a sequence that $x,y,z$ if ${y^2} = xz$ i.e., $y = \sqrt {xz} $ , it is called a geometric mean.
Properties of harmonic progression are , there are three terms in a sequence that $a,b,c$ if \[b = \dfrac{{2ab}}{{a + b}}\] , it is called the harmonic progression .

Complete step by step answer:
From the given data first we get $\dfrac{1}{{16}},a,b$ are in geometric progression
From properties of geometric progression , we get
${a^2} = b \times \dfrac{1}{{16}}$
$ \Rightarrow {a^2} = \dfrac{b}{{16}}$
Take cross multiplication of the above equation and get
$ \Rightarrow b = 16{a^2}$ …………………………………(1)
From the given data now we get $a,b,\dfrac{1}{6}$ are in harmonic progression .
From the properties of harmonic progression , we get
$b = \dfrac{{\left( {2 \times a \times \dfrac{1}{6}} \right)}}{{\left( {a + \dfrac{1}{6}} \right)}}$
$ \Rightarrow b = \dfrac{{\left( {\dfrac{a}{3}} \right)}}{{\left( {\dfrac{{6a + 1}}{6}} \right)}}$
$ \Rightarrow b = \dfrac{a}{{\left( {\dfrac{{6a + 1}}{2}} \right)}}$
$ \Rightarrow b = \dfrac{{2a}}{{6a + 1}}$
Take cross multiplication and we get
$ \Rightarrow b \times (6a + 1) = 2a$
Put the value of $b$ from the equation (1) in above equation and we have
$ \Rightarrow 16{a^2}(6a + 1) = 2a$
Divide both sides of the above equation by $a$ if $a \ne 0$
$ \Rightarrow 16a(6a + 1) = 2$
Multiplying and get
$ \Rightarrow 96{a^2} + 16a = 2$
$ \Rightarrow 96{a^2} + 16a - 2 = 0$
Divide both side of the above equation by $2$ and get
$ \Rightarrow 48{a^2} + 8a - 1 = 0$
Now we take middle term method to find the values of $a$
$ \Rightarrow 48{a^2} + (12 - 4)a - 1 = 0$
$ \Rightarrow 48{a^2} + 12a - 4a - 1 = 0$
Take common part by part and get
$ \Rightarrow 12a(4a + 1) - 1(4a + 1) = 0$
$ \Rightarrow (4a + 1)(12a - 1) = 0$
Therefore $4a + 1 = 0$ or $12a - 1 = 0$
$ \Rightarrow a = - \dfrac{1}{4}$ or $a = \dfrac{1}{{12}}$
Put $a = - \dfrac{1}{4}$ in (1) and we get
$b = 16 \times {\left( { - \dfrac{1}{4}} \right)^2}$
$ \Rightarrow b = 16 \times \dfrac{1}{{16}}$
$ \Rightarrow b = 1$
Again put the value of $a = \dfrac{1}{{12}}$ in (1) and get
$b = 16 \times {\left( {\dfrac{1}{{12}}} \right)^2}$
$ \Rightarrow b = 16 \times \dfrac{1}{{144}}$
$ \Rightarrow b = \dfrac{1}{9}$
$\therefore $ When $a = - \dfrac{1}{4}$ then $b = 1$ and $a = \dfrac{1}{{12}}$ then $b = \dfrac{1}{9}$. So, option (C) is correct.

Note:
 In quadratic factorization using the middle term which is the $a$ th term is the sum of the two factors and product equal to the last tem. If the product of two factors equals zero then either the first factor is zero or the second factor should be zero. i.e., any one of them should be zero. Squaring both positive or negative terms always gives positive terms.