To minimize the standard deviation (S.D.) of a list of numbers, the numbers in the list should be as close as possible to the mean of the list. In other words, the data points should be clustered tightly around the mean.

The original list of 5 terms is: $\(5,10,15,20,25\)$.
This is an arithmetic progression.
The mean of this original list is $\(\frac{5+10+15+20+25}{5}=\frac{75}{5}=15\)$.
The median of this original list is also 15.

When a 6th term $X$ is added to the list, we will have 6 terms.
Let the new list be $\(L^{\prime}=\{5,10,15,20,25, X\}\)$.
To minimize the standard deviation, the new term $X$ should be as close as possible to the mean of the new list, and it should also help in making all terms as close to the mean as possible.

Consider the properties of standard deviation:
Standard deviation measures the spread of data points around the mean.
If all data points are identical, the standard deviation is 0 (its minimum possible value).
If we add a new term $X$ to the list, the new mean will be $\(\frac{75+X}{6}\)$.
For the standard deviation to be minimum, the new term $X$ should be equal to the mean of the original list, which is 15.
Let's test this intuition. If $X=15$, the list becomes $\(\{5,10,15,15,20,25\}\)$.

Let's calculate the mean of this new list:
New Mean $\(=\frac{5+10+15+15+20+25}{6}=\frac{90}{6}=15\)$.
Notice that if $\(X=15\)$, the mean of the new list is still 15 .

Now, let's consider the spread around the mean (15) for $X=15$ :
The deviations from the mean are:

$$
\(\begin{aligned}
& 5-15=-10 \\
& 10-15=-5 \\
& 15-15=0 \\
& 15-15=0 \\
& 20-15=5 \\
& 25-15=10
\end{aligned}\)
$$


The sum of squared deviations is $\((-10)^2+(-5)^2+0^2+0^2+5^2+10^2=100+25+0+$ $0+25+100=250\)$.
The variance is $\(250 / 6=41.67\)$.
The standard deviation is $\(\sqrt{41.67} \approx 6.45\)$.
What if $X$ is a different value?
Suppose $X=14$.
New list: $\(\{5,10,14,15,20,25\}\)$.
New Mean $\(=\frac{75+14}{6}=\frac{89}{6} \approx 14.833\)$.
The deviations from the mean would be:

$$
\(\begin{aligned}
& 5-14.833=-9.833 \\
& 10-14.833=-4.833 \\
& 14-14.833=-0.833 \\
& 15-14.833=0.167 \\
& 20-14.833=5.167 \\
& 25-14.833=10.167
\end{aligned}\)
$$


The sum of squared deviations will be larger than for $X=15$ because the new mean is no longer an integer, and the values are generally further from it.
General Principle:
To minimize the standard deviation of a dataset, the new data point should be placed at the mean of the existing data, or as close to the center of the existing data as possible. If the new data point is equal to the mean of the existing data, it will not shift the mean, and it will add a deviation of 0 to the sum of squared deviations, thus minimizing the overall spread.

In this case, the mean of the original 5 terms is 15 . If we add $X=15$, the new list will have 15 as its mean, and the new term $X$ will have a deviation of 0 from the mean, which contributes nothing to the spread. Any other value for $X$ would either shift the mean away from 15 (making the other terms further from the mean), or if $X$ is not 15 , its own deviation from the mean will be non-zero, increasing the sum of squared deviations.

Therefore, to minimize the standard deviation, $X$ should be equal to the mean of the original list, which is 15.

The final answer is 15 .