The number of integral terms in the expansion of 3 1 2 + 5 1 4 680 is equal to

Given expansion is ${\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)}^{680}$.

The general term in the expansion ${\left(x+a\right)}^n$ is $T_{r+1}={}^{n}C_r·x^{n-r}·a^r$

$\Rightarrow T_{r+1}={}^{680}C_r{\left(3^{\frac{1}{2}}\right)}^{680-r}{\left(5^{\frac{1}{4}}\right)}^r, 0\le r\le 680$

$={}^{680}C_r{\left(3\right)}^{\frac{680-r}{2}}{\left(5\right)}^{\frac{r}{4}}$

For ${\left(3\right)}^{\frac{680-r}{2}}{\left(5\right)}^{\frac{r}{4}}$ to be rational $r$ should be a multiple of $4$.

$\Rightarrow r=0,4,8,12,.....680$

$\Rightarrow a_n=a+\left(n-1\right)d$

$\Rightarrow 680=0+\left(n-1\right)4$

$\Rightarrow n=171$

That means $r$ can take $171$ values.

Hence, the required answer is $171$.