Suppose that the mean and median of the non-negative numbers $$21, 8, 17, a, 51, 103, b, 13, 67$$, $$(a > b)$$, are 40 and 21, respectively. If the mean deviation about the median is 26, then $$2a$$ is equal to :

The nine numbers are $$21,\,8,\,17,\,a,\,51,\,103,\,b,\,13,\,67$$ with $$a \gt b \ge 0$$.

Step 1 : Mean condition
For $$n = 9$$ numbers, mean $$\mu = 40$$ gives $$\sum x_i = n\mu = 9 \times 40 = 360.$$ Sum of the seven known values is $$21+8+17+51+103+13+67 = 280.$$ Hence $$a + b = 360 - 280 = 80.$$ $$-(1)$$

Step 2 : Median condition
With nine values, the 5-th term of the ascending list is the median. Given median = $$21$$ and the number $$21$$ is already present, we must have exactly four values $$\le 21$$ and four values $$\ge 21$$ besides the median itself.
Among the known numbers, those $$\lt 21$$ are $$8,\,13,\,17$$ (3 numbers) and those $$\gt 21$$ are $$51,\,67,\,103$$ (3 numbers). To reach the count 4-and-4, one of $$a,\,b$$ must be $$\lt 21$$ and the other $$\gt 21$$. Because $$a \gt b$$, we conclude

$$b \lt 21,\qquad a \gt 21.$$ $$-(2)$$

Step 3 : Mean deviation about the median
Mean deviation about median $$M$$ is $$\text{MD} = \frac{1}{n}\sum_{i=1}^{n}\lvert x_i - M\rvert.$$ Here $$M = 21$$ and MD is given as $$26$$, so $$\sum_{i=1}^{9}\lvert x_i - 21\rvert = 9 \times 26 = 234.$$

Compute the absolute deviations of the known numbers:
$$\begin{aligned} |21-21| &= 0,\\ |8-21| &= 13,\\ |17-21| &= 4,\\ |51-21| &= 30,\\ |103-21|&= 82,\\ |13-21| &= 8,\\ |67-21| &= 46. \end{aligned}$$ Their sum is $$0+13+4+30+82+8+46 = 183.$$

Let the remaining two deviations be $$|a-21| + |b-21|$$. They must satisfy $$|a-21| + |b-21| = 234 - 183 = 51.$$ $$-(3)$$

Using (2), $$|a-21| = a-21,\qquad |b-21| = 21-b,$$ so (3) becomes $$(a-21) + (21 - b) = a - b = 51.$$ $$-(4)$$

Step 4 : Solve for $$a$$ and $$b$$
From (1) and (4): $$a + b = 80,$$ $$a - b = 51.$$ Adding, $$2a = 80 + 51 = 131.$$ Thus $$2a = 131.$$

Therefore, the required value is
Option D which is: 131