If a,b,c,d are positive real numbers such that `(a)/(3) = (a+b)/(4)= (a+b+c)/(5) = (a+b+c+d)/(6)`, then `(a)/(b+2c+3d)` is:-

Options

A. `1//2`
B. 1
C. 2
D. Not determinable

Answer

To solve the problem, we start with the given equations:

\[ \frac{a}{3} = \frac{a+b}{4} = \frac{a+b+c}{5} = \frac{a+b+c+d}{6} \]

Let's denote the common value of these fractions as \( k \). Thus, we can write:

1. \( \frac{a}{3} = k \)
2. \( \frac{a+b}{4} = k \)
3. \( \frac{a+b+c}{5} = k \)
4. \( \frac{a+b+c+d}{6} = k \)

From these equations, we can express \( a \), \( b \), \( c \), and \( d \) in terms of \( k \):

### Step 1: Express \( a \) in terms of \( k \)
From the first equation, we have:
\[ a = 3k \]

### Step 2: Express \( b \) in terms of \( k \)
From the second equation:
\[ a + b = 4k \implies 3k + b = 4k \implies b = 4k - 3k = k \]

### Step 3: Express \( c \) in terms of \( k \)
From the third equation:
\[ a + b + c = 5k \implies 3k + k + c = 5k \implies c = 5k - 4k = k \]

### Step 4: Express \( d \) in terms of \( k \)
From the fourth equation:
\[ a + b + c + d = 6k \implies 3k + k + k + d = 6k \implies d = 6k - 5k = k \]

### Summary of Values
Now we have:
- \( a = 3k \)
- \( b = k \)
- \( c = k \)
- \( d = k \)

### Step 5: Calculate \( b + 2c + 3d \)
Now, we need to find \( b + 2c + 3d \):
\[ b + 2c + 3d = k + 2(k) + 3(k) = k + 2k + 3k = 6k \]

### Step 6: Calculate \( \frac{a}{b + 2c + 3d} \)
Now we can find \( \frac{a}{b + 2c + 3d} \):
\[ \frac{a}{b + 2c + 3d} = \frac{3k}{6k} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
Thus, the value of \( \frac{a}{b + 2c + 3d} \) is:
\[ \frac{1}{2} \]