Questions & Answers

Question

Answers

A). 100

B). 50

C). 101

D). None of these

Answer
Verified

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