Average Calculation: Finding $a_4$

The problem asks for the value of $a_4$ given the average of four numbers ($a_1, a_2, a_3, a_4$), the value of $a_1$, and a condition relating the averages of two sets of three numbers.

Given Information

• Average of $a_1, a_2, a_3, a_4$ is $19.5$.
• $a_1 = 21$.
• Average of $a_1, a_2, a_3$ = Average of $a_2, a_3, a_4$.

Step-by-Step Solution

1. Set up the equation for the average of four numbers:

   The average of $a_1, a_2, a_3$, and $a_4$ is given by:

   $ \frac{a_1 + a_2 + a_3 + a_4}{4} = 19.5 $

   Multiply both sides by 4:

   $ a_1 + a_2 + a_3 + a_4 = 19.5 \times 4 $

   $ a_1 + a_2 + a_3 + a_4 = 78 \quad \quad (1) $

2. Use the condition relating the averages of subsets:

   The average of $a_1, a_2, a_3$ is:

   $ \frac{a_1 + a_2 + a_3}{3} $

   The average of $a_2, a_3, a_4$ is:

   $ \frac{a_2 + a_3 + a_4}{3} $

   According to the problem statement, these averages are equal:

   $ \frac{a_1 + a_2 + a_3}{3} = \frac{a_2 + a_3 + a_4}{3} $

3. Simplify the condition:

   Multiply both sides by 3:

   $ a_1 + a_2 + a_3 = a_2 + a_3 + a_4 $

   Subtract $a_2$ and $a_3$ from both sides:

   $ a_1 = a_4 $

4. Find the value of $a_4$:

   We are given $a_1 = 21$. Since we found $a_1 = a_4$, we can directly conclude:

   $ a_4 = 21 $

   This result is consistent with Equation (1). Substituting $a_1 = 21$ and $a_4 = 21$ into Equation (1) gives $21 + a_2 + a_3 + 21 = 78$, simplifying to $a_2 + a_3 = 36$. The condition $a_1 = a_4$ is sufficient to find $a_4$.

Therefore, the value of $a_4$ is 21.