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**Hint:**Here we can write the general binomial expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$ which we can write by:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$\[ = \sum\limits_{r = 0}^n {{}^n{C_r}} {({x^{\dfrac{1}{2}}})^{n - r}}{\left( {\dfrac{1}{{{2^{\dfrac{1}{4}}}}}} \right)^r}\]

And then we can write the coefficients of the first three terms as $2b = a + c$ if the first three terms are $a,b,c$ and then we can find the number of terms easily with integral powers of $x$.

**Complete step by step solution:**

Here we are given to write the expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$

So we must know that the general binomial expansion is given by:

${\left( {a + b} \right)^n}$\[ = \sum\limits_{r = 0}^n {{}^n{C_r}} {(a)^{n - r}}{\left( b \right)^r}\]

So similarly we can write the expansion of the given binomial expansion and write it as:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$\[ = \sum\limits_{r = 0}^n {{}^n{C_r}} {({x^{\dfrac{1}{2}}})^{n - r}}{\left( {\dfrac{1}{{2{x^{\dfrac{1}{4}}}}}} \right)^r}\]

Now we can simplify and write it as:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$\[ = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{\dfrac{{n - r}}{2}}}{\left( {\dfrac{1}{2}} \right)^r}{x^{\dfrac{{ - r}}{4}}}\]$ = \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{\dfrac{{n - r}}{2} - \dfrac{r}{4}}}$

So we get the simplified form as:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$ = \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{\dfrac{{2n - 3r}}{4}}}$$ - - - - - (1)$

Now we have to expand the decreasing power of $x$ so we can substitute the values of $r = 0,1,2,....$ and so on.

So we get:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$ = {}^n{C_0}{\left( {\dfrac{1}{2}} \right)^0}{(x)^{\dfrac{{2n - 3\left( 0 \right)}}{4}}} + {}^n{C_1}{\left( {\dfrac{1}{2}} \right)^1}{(x)^{\dfrac{{2n - 3\left( 1 \right)}}{4}}} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}{(x)^{\dfrac{{2n - 3\left( 2 \right)}}{4}}} + .........$

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$ = {}^n{C_0}{(x)^{\dfrac{n}{2}}} + {}^n{C_1}\left( {\dfrac{1}{2}} \right){(x)^{\dfrac{{2n - 3}}{4}}} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}{(x)^{\dfrac{{2n - 6}}{4}}} + .........$$ - - - - - \left( 2 \right)$

So we get this as the expansion of the term in the decreasing powers of the variable $x$

Now we are also given that the first three coefficients are in arithmetic progression which means these three terms have the common difference between them.

We know that if $a,b,c$ are in AP then $2b = a + c$

Hence we can write from the expansion of equation $\left( 2 \right)$ that:

\[{}^n{C_0},{}^n{C_1}\left( {\dfrac{1}{2}} \right),{}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}\] are in AP, hence we can write that:

$

2{}^n{C_1}\left( {\dfrac{1}{2}} \right) = {}^n{C_0} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2} \\

{}^n{C_1} = 1 + \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}.\dfrac{1}{4} \\

$

${}^n{C_1} = 1 + \dfrac{{n\left( {n - 1} \right)}}{8}$

Hence we can write that:

$

n = 1 + \dfrac{{n\left( {n - 1} \right)}}{8} \\

\left( {n - 1} \right) - \dfrac{{n\left( {n - 1} \right)}}{8} = 0 \\

\left( {n - 1} \right)\left( {1 - \dfrac{n}{8}} \right) = 0 \\

$

We can further write it as:

$

- \dfrac{1}{8}\left( {n - 1} \right)\left( {n - 8} \right) = 0 \\

\left( {n - 1} \right)\left( {n - 8} \right) = 0 \\

n = 1{\text{ or 8}} \\

$

But we know that $n$ cannot be $1$ because we are told that there are at least $3$ terms. Hence we can say that $n = 8$

Substitute $n = 8$ in equation (1) we get:

${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^8}$$ = \sum\limits_{r = 0}^8 {{}^8{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{4 - \dfrac{{3r}}{4}}}$

Hence know that power of the $x$ has to be integer which means that $4 - \dfrac{{3r}}{4} \in Z$

Therefore $\dfrac{{3r}}{4} \in Z$

So $r$ has to be multiple of $4$ so that denominator and numerator cancel each other and we get the integer.

So as $r \in \left[ {0,8} \right]$

Hence we can say that $r = 0,4,8$

**Hence we can say that there are $3$ values of $r$**

Option C) is the correct answer.

Option C) is the correct answer.

**Note:**

Whenever the student is given to expand the terms of the bracket of the form ${\left( {a + b} \right)^n}$ then we must know that its expansion can be written in the form of ${\left( {a + b} \right)^n}$\[ = \sum\limits_{r = 0}^n {{}^n{C_r}} {(a)^{n - r}}{\left( b \right)^r}\]

Therefore according to this expansion we can solve for what we are required to find.

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