SEQUENCES AND SERIES - 3 (Harmonic Progression)

A sequence $a _{1}, a _{2} \ldots \ldots . a _{n}$ of non-zero numbers is called a harmonic sequence if the sequence $\frac{1}{a _{1}}, \frac{1}{a _{2}}, \ldots \ldots . . \frac{1}{a _{n}}, \ldots \ldots .$. is an A.P.

Eg: The sequence $\frac{1}{3}, \frac{1}{5}, \frac{1}{7} \ldots .$. is an H.P. because the sequence $3,5,7 \ldots \ldots$ is an A.P.

$n^{\text {th }}$ term of a H.P.

The $\mathrm{n}^{\text {th }}$ term of a H.P. is the reciprocal of $\mathrm{n}^{\text {th }}$ term of the corresponding A.P. and the common difference of the corresponding A.P is (d). i.e. $d=\frac{1}{a _{2}}-\frac{1}{a _{1}}$.

i. $\mathrm{n}^{\text {th }}$ term of the H.P is given by

$ a _{n}=\frac{1}{\frac{1}{a _{1}}+(n-1)(d)}=\frac{1}{\frac{1}{a _{1}}+(n-1)\left(\frac{1}{a _{2}}-\frac{1}{a _{1}}\right)}=\frac{a _{1} a _{2}}{a _{2}+(n-1)\left(a _{1}-a _{2}\right)} $

ii. $\mathrm{n}^{\text {th }}$ term of the H.P. from end

$ \begin{aligned} & \mathrm{a} _{\mathrm{n}}^{1}=\frac{1}{\frac{1}{\mathrm{a} _{\mathrm{n}}}+(\mathrm{n}-1)(-\mathrm{d})}=\frac{1}{\frac{1}{\mathrm{a} _{\mathrm{n}}}-(\mathrm{n}-1)\left(\frac{1}{\mathrm{a} _{2}}-\frac{1}{\mathrm{a} _{1}}\right)} \\ & =\frac{\mathrm{a} _{1} \mathrm{a} _{2} \mathrm{a} _{\mathrm{n}}}{\mathrm{a} _{1} \mathrm{a} _{2}-(\mathrm{n}-1) \mathrm{a} _{\mathrm{n}}\left(\mathrm{a} _{1}-\mathrm{a} _{2}\right)} \end{aligned} $

Note: No term of H.P. can be zero and there is no general formula for finding out the sum of $n$ terms of a H.P.

Harmonic Mean

If $a, b, c$ are in H.P., then $b$ is called the H.M. between $a \& c$.

Now a,b,c are in H.P.

$ \begin{aligned} & \Rightarrow \quad \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text { are is A.P. } \\ & \Rightarrow \quad \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \\ & \Rightarrow \quad b=\frac{2 a c}{a+c} \end{aligned} $

i.e. H.M. between $\mathrm{a} \& \mathrm{c}$ is $\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}$

Note: The single H.M. between $a \& b$ is $\frac{2 a b}{a+b}$.

The single H.M. (H) of $n$ positive numbers $\mathrm{a} _{1}, \mathrm{a} _{2}, \ldots . \mathrm{a} _{\mathrm{n}}$ is given by

$\frac{1}{\mathrm{H}}=\frac{\frac{1}{\mathrm{a} _{1}}+\frac{1}{\mathrm{a} _{2}}+\frac{1}{\mathrm{a} _{3}}+\ldots \ldots .+\frac{1}{\mathrm{a} _{\mathrm{n}}}}{\mathrm{n}}=\frac{1}{\mathrm{n}}\left(\frac{1}{\mathrm{a} _{1}}+\frac{1}{\mathrm{a} _{2}}+\frac{1}{\mathrm{a} _{3}}+\ldots \ldots .+\frac{1}{\mathrm{a} _{\mathrm{n}}}\right)$

Insertion of Harmonic Means

Let $\mathrm{a}$ and $\mathrm{b}$ be two given numbers and $\mathrm{H} _{1}, \mathrm{H} _{2} \ldots . . \mathrm{H} _{\mathrm{n}}$ be the H.M.’s between them. Then $\mathrm{a}$, $\mathrm{H} _{1}, \mathrm{H} _{2} \ldots \ldots . \mathrm{H} _{\mathrm{n}}$, b will be in H.P. Let $\mathrm{d}$ the common differece of the corresponding A.P. $\mathrm{b}=(\mathrm{n}+2)^{\text {th }}$ term of H.P.

$\mathrm{b}=\frac{1}{\frac{1}{\mathrm{a}}+(\mathrm{n}+2-1) \mathrm{d}}$

$\frac{1}{b}=\frac{1}{a}+(n+1) d \Rightarrow d=\frac{\frac{1}{b}-\frac{1}{a}}{n+1}$

$d=\frac{a-b}{(n+1) a b}$

$\therefore \quad \frac{1}{\mathrm{H} _{1}}=\frac{1}{\mathrm{a}}+\mathrm{d}=\frac{1}{\mathrm{a}}+\frac{\mathrm{a}-\mathrm{b}}{(\mathrm{n}+1) \mathrm{ab}}$

$\frac{1}{\mathrm{H} _{2}}=\frac{1}{\mathrm{a}}+2 \mathrm{~d}=\frac{1}{\mathrm{a}}+\frac{2(\mathrm{a}-\mathrm{b})}{(\mathrm{n}+1) \mathrm{ab}}$

$\frac{1}{\mathrm{H} _{\mathrm{n}}}=\frac{1}{\mathrm{a}}+\mathrm{nd}=\frac{1}{\mathrm{a}}+\frac{\mathrm{n}(\mathrm{a}-\mathrm{b})}{(\mathrm{n}+1) \mathrm{ab}}$

E.g: Insert 4 H.M’s between $\frac{2}{3}$ and $\frac{2}{13}$

$ \begin{aligned} & \mathrm{d}=\frac{\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{a}}}{\mathrm{n}+1}=\frac{\frac{13}{2}-\frac{3}{2}}{4+1} \\ & \mathrm{~d}=1 \\ & \therefore \quad \frac{1}{\mathrm{H} _{1}}=\frac{3}{2}+1=\frac{5}{2} \text { or } \mathrm{H} _{1}=\frac{2}{5} \end{aligned} $

$ \begin{aligned} & \frac{1}{\mathrm{H} _{2}}=\frac{3}{2}+2=\frac{7}{2} \text { or } \mathrm{H} _{2}=\frac{2}{7} \\ & \frac{1}{\mathrm{H} _{3}}=\frac{3}{2}+3=\frac{9}{2} \text { or } \mathrm{H} _{3}=\frac{2}{9} \\ & \frac{1}{\mathrm{H} _{4}}=\frac{3}{2}+4=\frac{11}{2} \text { or } \mathrm{H} _{4}=\frac{2}{11} \end{aligned} $

Note: The sum of reciprocals of $n$ H.M’s between two numbers is $n$ times the reciprocal of single H.M. between them.

i.e. if a,b are the numbers and $\mathrm{H} _{1}, \mathrm{H} _{2} \ldots . . \mathrm{H} _{\mathrm{n}}$ be the H.M’s between them,

$ \begin{aligned} & \text { then } \frac{1}{\mathrm{H} _{1}}+\frac{1}{\mathrm{H} _{2}}+\ldots \ldots .+\frac{1}{\mathrm{H} _{\mathrm{n}}}=\mathrm{n}\left(\frac{\mathrm{a}+\mathrm{b}}{2 \mathrm{ab}}\right) \\ & =\mathrm{n}\left(\frac{1}{\text { H.M. of } \mathrm{a} \text { and } \mathrm{b}}\right) \end{aligned} $

Solved Examples

1. If $\mathrm{a} _{1}, \mathrm{a} _{2}, \mathrm{a} _{3} \ldots \ldots \mathrm{a} _{10}$ are in A.P. and $\mathrm{h} _{1}, \mathrm{~h} _{2}, \mathrm{~h} _{3}, \ldots . . \mathrm{h} _{10}$ are in H.P. If $\mathrm{a} _{1}=\mathrm{h} _{1}=2$ and $\mathrm{a} _{10}=\mathrm{h} _{10}=3$, then $\mathrm{a} _{4} \mathrm{~h} _{7}$ is

(a). 6

(b). 7

(c). $18$

(d). none of these

Show Answer

Solution:

$a _{10}=3 \Rightarrow 3=2+9 d \Rightarrow d=\frac{1}{9}$

$\mathrm{h} _{1}, \mathrm{~h} _{2} \ldots . . \mathrm{h} _{10}$ are in H.P.

$\frac{1}{\mathrm{~h} _{10}}=\frac{1}{\mathrm{~h} _{1}}+9 \mathrm{~d} _{1} \Rightarrow \frac{1}{3}=\frac{1}{2}+9 \mathrm{~d} _{1} \Rightarrow \mathrm{d} _{1}=\frac{-1}{54}$

$\mathrm{a} _{4}=\mathrm{a} _{1}+3 \mathrm{~d}=2+3 \mathrm{x} \frac{1}{9}=2+\frac{1}{3}=\frac{7}{3}$

and $\frac{1}{\mathrm{~h} _{7}}=\frac{1}{\mathrm{~h} _{1}}+6 \mathrm{~d} \Rightarrow \mathrm{h} _{7}=\frac{18}{7}$

$\therefore \quad \mathrm{a} _{4} \mathrm{~h} _{7}=\frac{7}{3} \times \frac{18}{7}=6$

Answer: a

2. If $a _{1}, a _{2}, a _{3} \ldots . a _{n}$ are in H.P., then $a _{1} a _{2}+a _{2} a _{3}+\ldots \ldots .+a _{n-1} a _{n}$ will be equal

(a). $a_1 a_n$

(b). $na_1 a_n$

(c). $(n-1) a_1 a_n$

(d). none of these

Show Answer

Solution:

$\frac{1}{\mathrm{a} _{1}}, \frac{1}{\mathrm{a} _{2}}, \ldots \ldots \ldots \ldots \frac{1}{\mathrm{a} _{\mathrm{n}}}$ are in A.P.

$\Rightarrow \quad \frac{1}{a _{2}}-\frac{1}{a _{1}}=\frac{1}{a _{3}}-\frac{1}{a _{2}}=\ldots \ldots . \frac{1}{a _{n}}-\frac{1}{a _{n-1}}=d$

$ \begin{aligned} & \Rightarrow \quad \frac{a _{1}-a _{2}}{a _{1} a _{2}}=\frac{a _{2}-a _{3}}{a _{2} a _{3}}=\ldots \ldots .=\frac{a _{n-1}-a _{n}}{a _{n} a _{n-1}}=d \\ & \Rightarrow \quad a _{1}-a _{2}=d a _{1} a _{2} \\ & \Rightarrow \quad a _{2}-a _{3}=d a _{2} a _{3} \\ & \Rightarrow \quad a _{n-1}-a _{n}=d a _{n} a _{n-1} \end{aligned} $

Adding, we get $\mathrm{a} _{1}-\mathrm{a} _{\mathrm{n}}=d\left(\mathrm{a} _{1} \mathrm{a} _{2}+\mathrm{a} _{2} \mathrm{a} _{3}+\ldots \ldots+\mathrm{a} _{\mathrm{n}} \mathrm{a} _{\mathrm{n}-1}\right)$

i.e. $a _{1} a _{2}+a _{2} a _{3}+\ldots \ldots+a _{n} a _{n-1}=\frac{a _{1}-a _{n}}{d}$

$=\frac{\left(a _{1}-a _{n}\right)}{\left(a _{1}-a _{n}\right)} a _{1} a _{n}(n-1)$

$=\mathrm{a} _{1} \mathrm{a} _{\mathrm{n}}(\mathrm{n}-1)$

$\left(\begin{array}{c}\frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\ \Rightarrow \frac{1}{a_n}-\frac{1}{a_1}=(n-1) d \\ \frac{a_1-a_n}{d}=(n-1) a_1 a n\end{array}\right)$

Answer: c

3. If $\mathrm{A} _{1}, \mathrm{~A} _{2} ; \mathrm{G} _{1}, \mathrm{G} _{2}$ and $\mathrm{H} _{1}, \mathrm{H} _{2}$ be two A.M.’s , G.M.’s and H.M.’s between two numbers a & b repectively, then $\frac{\mathrm{G} _{1} \mathrm{G} _{2}}{\mathrm{H} _{1} \mathrm{H} _{2}} \times \frac{\mathrm{H} _{1}+\mathrm{H} _{2}}{\mathrm{~A} _{1}+\mathrm{A} _{2}}=$

(a). 1

(b). 0

(c). 2

(d). 3

Show Answer

Solution:

$\mathrm{a}, \mathrm{A} _{1}, \mathrm{~A} _{2}, \mathrm{~b}$ are in A.P $\Rightarrow \quad \mathrm{A} _{1}+\mathrm{A} _{2}=\mathrm{a}+\mathrm{b}$

$\mathrm{a}, \mathrm{G} _{1}, \mathrm{G} _{2}, \mathrm{~b}$ are in G.P $\Rightarrow \mathrm{G} _{1} \mathrm{G} _{2}=\mathrm{ab}$

$\mathrm{a}, \mathrm{H} _{1}, \mathrm{H} _{2}, \mathrm{~b}$ are in H.P $\Rightarrow \quad \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{H} _{1}}, \frac{1}{\mathrm{H} _{2}}, \frac{1}{\mathrm{~b}}$ are is A.P

$\Rightarrow \quad \frac{1}{\mathrm{H} _{1}}+\frac{1}{\mathrm{H} _{2}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}$

Given expression is $\frac{\mathrm{G} _{1} \mathrm{G} _{2}}{\mathrm{~A} _{1}+\mathrm{A} _{2}}\left(\frac{1}{\mathrm{H} _{1}}+\frac{1}{\mathrm{H} _{2}}\right)=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}\right)=1$

Answer: a

4. If nine A.M.’s and nine H.M.’s are inserted between $2 \& 3$, then $A+\frac{6}{H}=\ldots$ (where $A$ is any of the A.M.’s and $\mathrm{H}$ the corresponding H.M.)

(a). 5

(b). 3

(c). 15

(d). none of these

Show Answer

Solution:

$2, \mathrm{~A} _{1}, \mathrm{~A} _{2} \ldots . . \mathrm{A} _{9}, 3$ are in A.P.

$\mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}+1}=\frac{1}{10}$

$A _{i}=2+i d=2+\frac{i}{10} ; i=1,2, \ldots \ldots 9$

$ \begin{aligned} & \frac{1}{2}, \frac{1}{\mathrm{H} _{1}}, \frac{1}{\mathrm{H} _{2}}, \ldots \ldots \ldots \frac{1}{\mathrm{H} _{9}}, \frac{1}{3} \text { are in A.P. } \\ & \mathrm{D}=\frac{\mathrm{a}-\mathrm{b}}{(\mathrm{n}+1) \mathrm{ab}}=\frac{-1}{60} ; \mathrm{i}=1,2, \ldots \ldots .9 \\ & \frac{1}{\mathrm{H} _{\mathrm{i}}}=\frac{1}{2}+\mathrm{iD}=\frac{1}{2}+\mathrm{i}\left(\frac{-1}{60}\right) ; \mathrm{i}=1,2, \ldots \ldots, 9 \\ & \frac{6}{\mathrm{H} _{\mathrm{i}}}=3-\frac{\mathrm{i}}{10} ; \mathrm{i}=1,2, \ldots \ldots \ldots .9 \\ & \therefore \quad \mathrm{A} _{\mathrm{i}}+\frac{6}{\mathrm{H} _{\mathrm{i}}}=5 ; \mathrm{i}=1,2,3 \ldots \ldots \ldots .9 \end{aligned} $

Answer: a

5. If $\mathrm{H} _{1}, \mathrm{H} _{2} \ldots . . \mathrm{H} _{20}$ be 20 harmonic means between $2 \& 3$, then $\frac{\mathrm{H} _{1}+2}{\mathrm{H} _{1}-2}+\frac{\mathrm{H} _{20}+3}{\mathrm{H} _{20}-3}=$

(a). 20

(b). 21

(c). 40

(d). 38

Show Answer

Solution:

$\frac{\mathrm{H} _{1}+2}{\mathrm{H} _{1}-2}+\frac{\mathrm{H} _{20}+3}{\mathrm{H} _{20}-3}$

$=\frac{\frac{1}{2}+\frac{1}{\mathrm H_1}}{\frac{1}{2}-\frac{1}{\mathrm H_1}}+\frac{\frac{1}{3}+\frac{1}{\mathrm H_{20}}}{\frac{1}{3}-\frac{1}{\mathrm{H}_{20}}}$

$=\frac{\frac{1}{2}+\frac{1}{2}+\mathrm{d}}{\frac{1}{2}-\frac{1}{2}-\mathrm{d}}+\frac{\frac{1}{3}+\frac{1}{3}-\mathrm{d}}{\frac{1}{3}-\frac{1}{3}+\mathrm{d}}$

$=\frac{1+\mathrm{d}}{-\mathrm{d}}+\frac{\frac{2}{3}-\mathrm{d}}{\mathrm{d}}=\frac{-1}{\mathrm{~d}}-1+\frac{2}{3 \mathrm{~d}}-1$

$=\frac{1}{\mathrm{~d}}\left(\frac{2}{3}-1\right)-2$

$ =\frac{1}{\mathrm{~d}} \times 42 \mathrm{~d}-2 \quad\left(\because \frac{1}{3}=\frac{1}{2}+(22-1) \mathrm{d} \Rightarrow \frac{2}{3}-1=42 \mathrm{~d}\right) $

Answer: c

Practice questions

1. If $x, y, z$ are in H.P., then the value of expression $\log (x+z)+\log (x-2 y+z)$ will be

(a). $d\log (\mathrm{x}-\mathrm{z})$

(b). $d2 \log (\mathrm{x}-\mathrm{z})$

(c). $d3 \log (\mathrm{x}-\mathrm{z})$

(d). $d4 \log (\mathrm{x}-\mathrm{z})$

Show Answer Answer: (b)

2. If $a, b, c, d$ are positive real numbers such that $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=2$, then $\mathrm{M}=(\mathrm{a}+\mathrm{b})(\mathrm{c}+\mathrm{d})$ satisfies the equation

(a). $0<\mathrm{M} \leq 1$

(b). $d1 \leq \mathrm{M} \leq 2$

(c). $d2 \leq \mathrm{M} \leq 3$

(d). $3 \leq \mathrm{M} \leq 4$

Show Answer Answer: (a)

3. If $a _{1}, a _{2}, a _{3} \ldots \ldots, a _{n}$ are in H.P., then $\frac{a _{1}}{a _{2}+a _{3} \ldots .+a _{n}}, \frac{a _{2}}{a _{1}+a _{3} \ldots .+a _{n}}, \ldots \ldots \ldots ., \frac{a _{n}}{a _{1}+a _{2} \ldots .+a _{n-1}}$ are in.

(a). A.P.

(b). G.P.

(c). H.P.

(d). none of these

Show Answer Answer: (c)

4. If $\mathrm{a}, \mathrm{a} _{1}, \mathrm{a} _{2} \ldots \ldots \mathrm{a} _{2 \mathrm{n-1}}, \mathrm{b}$ are in A.P. ; $\mathrm{a}, \mathrm{b} _{1}, \mathrm{~b} _{2} \ldots \ldots \mathrm{b} _{2 \mathrm{n-1}-1}, \mathrm{~b}$ are in G.P. ; $\mathrm{a} _{1} \mathrm{c} _{1}, \mathrm{c} _{2} \ldots \ldots \mathrm{c} _{2 \mathrm{n}-1}, \mathrm{~b}$ are in H.P. where $a, b$, are positive, then the equation $a _{n} x^{2}-b _{n} x+c _{n}=0$ has its roots

(a). real and unequal

(b). real and equal

(c). imaginary

(d). none of these

Show Answer Answer: (c)

5. If $\mathrm{b}-\mathrm{c}, 2 \mathrm{~b}-\mathrm{x}, \mathrm{b}-\mathrm{a}$ are in H.P, then $\mathrm{a}-\frac{\mathrm{x}}{2}, \mathrm{~b}-\frac{\mathrm{x}}{2}$ and $\mathrm{c}-\frac{\mathrm{x}}{2}$ are in

(a). A.P

(b). G.P

(c). H.P

(d). none of these

Show Answer Answer: (b)

6. The first two terms of a H.P are $\frac{2}{5}$ and $\frac{12}{13}$ respectively. Then the largest term is

(a). $2^{\text {nd }}$ term

(b). $6^{\text {th }}$ term

(c). $4^{\text {th }}$ term

(d). none of these

Show Answer Answer: (a)

7. If a,b,c are in A.P, p,q,r are in H.P and ap, bq, cr are in G.P. then $\frac{\mathrm{p}}{\mathrm{r}}+\frac{\mathrm{r}}{\mathrm{p}}$ is equal to

(a). $ \frac{\mathrm{a}}{\mathrm{c}}-\frac{\mathrm{c}}{\mathrm{a}}$

(b). $ \frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}$

(c). $ \frac{\mathrm{b}}{\mathrm{q}}+\frac{\mathrm{q}}{\mathrm{b}}$

(d). $ \frac{b}{q}-\frac{q}{b}$

Show Answer Answer: (b)

8. If $a, b, c$ are in H.P, then the value of $\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{a}-\frac{1}{b}\right)$ is

(a). $\frac{2}{\mathrm{bc}}-\frac{1}{\mathrm{~b}^{2}}$

(b). $ \frac{1}{4}\left(\frac{3}{\mathrm{c}^{2}}+\frac{2}{\mathrm{ca}}-\frac{1}{\mathrm{c}^{2}}\right)$

(c). $\frac{3}{\mathrm{~b}^{2}}-\frac{2}{\mathrm{ab}}$

(d). none of these

Show Answer Answer: (a, c)

9. If $\mathrm{H} _{1}, \mathrm{H} _{2} \ldots \ldots . \mathrm{H} _{\mathrm{n}}$ be $\mathrm{n}$ harmonic means between $\mathrm{a}$ and $\mathrm{b}$, then $\frac{\mathrm{H} _{1}+\mathrm{a}}{\mathrm{H} _{1}-\mathrm{a}}+\frac{\mathrm{H} _{n}+\mathrm{b}}{\mathrm{H} _{\mathrm{n}}-\mathrm{b}}$ is equal to

(a). $0$

(b). $n$

(c). $2 \mathrm{n}$

(d). $1$

Show Answer Answer: (c)

10. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be in G.P. and $\mathrm{a}+\mathrm{x}, \mathrm{b}+\mathrm{x}, \mathrm{c}+\mathrm{x}$ in H.P., then the value of $\mathrm{x}$ is ( $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are district numbers)

(a). (c).

(b). $ \mathrm{b}$

(c). $ \mathrm{a}$

(d). none of these

Show Answer Answer: (b)

11. The harmonic mean of two numbers is 4, their A.M.A and G.M.G satisfy the relation $2 A+G^{2}=27$. The numbers are

(a). 6 and 3

(b). 3 and 6

(c). can’not find

(d). none of these

Show Answer Answer: (a, b)

12. If $\mathrm{n}$ be a root of the equation

$\mathrm{x}^{2}(1-\mathrm{ab})-\mathrm{x}\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)-(1+\mathrm{ab})=0$, then $\mathrm{H} _{1}-\mathrm{H} _{\mathrm{n}}=$

(a). $a b(a-b)$

(b). $a b(b-a)$

(c). $\frac{\mathrm{b}-\mathrm{a}}{\mathrm{ab}}$ (d). none of these

Show Answer Answer: (a)

13. If $x, y, z$ are in A.P, ax,by,cz in G.P. and $a, b, c$ in H.P., then $\frac{x}{z}+\frac{z}{x}=$

(a). $2$

(b). $ \frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}$

(c). $\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}$

(d). none of these

Show Answer Answer: (b)

14. If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in H.P., then the value of $\left(\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~d}^{2}}\right)\left(\frac{1}{\mathrm{~b}^{2}}-\frac{1}{\mathrm{c}^{2}}\right)$ is

(a). 1

(b). 2

(c). 3

(d). 4

Show Answer Answer: (c)


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