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**Hint:**The general term of expansion is given by \[{T_{r + 1}}\], so find the general term of the expansion then see that for what possible terms it can have integral coefficients, you will find an AP in those terms then use the formula for \[{n^{th}}\] to get the final results.

**Complete step-by-step answer:**We know that general term of any expansion given by \[{(a + b)^n}\] is \[{}^n{C_r} \times {a^{n - r}} \times {b^r}\]

So the general term \[{T_{r + 1}}\] in the expansion given in the question will be

\[\begin{array}{l}

{}^{600}{C_r}{\left( {{7^{\dfrac{1}{3}}}} \right)^{600 - r}} \times {\left( {{5^{\dfrac{1}{2}}}.x} \right)^r}\\

= {}^{600}{C_r} \times {7^{\left( {600 - r} \right) \times \dfrac{1}{3}}} \times {5^{\dfrac{r}{2}}} \times {x^r}\\

= {}^{600}{C_r} \times {7^{200 - \dfrac{r}{3}}} \times {5^{\dfrac{r}{2}}} \times {x^r}

\end{array}\]

Now \[{T_{r + 1}}\] is an integer if and only if \[\dfrac{r}{2}\& \dfrac{r}{3}\] are integers for all \[0 \le r \le 600\]

Thus we have all the multiples of 6 that will satisfy the condition

\[\therefore r = 0,6,12,......600\]

Since It is an AP

We know that the \[{n^{th}}\] term of an AP is given by \[{a_n} = a + (n - 1)d\]

Where \[{a_n}\] is the \[{n^{th}}\] term a is the first term and d is the common difference

Clearly here

\[\begin{array}{l}

a = 0\\

{a_n} = 600\\

d = 6

\end{array}\]

Which means that

\[\begin{array}{l}

\therefore {a_n} = a + (n - 1)d\\

\Rightarrow 600 = 0 + (n - 1) \times 6\\

\Rightarrow \dfrac{{600}}{6} = n - 1\\

\Rightarrow 100 = n - 1\\

\Rightarrow n = 101

\end{array}\]

Hence there are a total of 101 terms as integral coefficients

**Which means that C is the correct option here.**

**Note:**In exponential if we find \[{a^{{x^y}}}\] then it clearly means that \[{a^{x \times y}} = {a^{xy}}\] I have used this property of exponential while solving this question. Also, note that realizing that the possible values of r is in an AP was the key step which leads to the correct answer.

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