If a/b=b/c=c/d=d/e=e/f=1/3, then find the value of `((a+b+c+d+e)/(b+c+d+e+f))` :

Options

A. 1/81
B. 1/27
C. 0.33333333333333
D. 1

Answer

To solve the problem step by step, we start with the given ratios:

Given:
\[ \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e} = \frac{e}{f} = \frac{1}{3} \]

This means:
1. \( a = \frac{1}{3}b \)
2. \( b = \frac{1}{3}c \)
3. \( c = \frac{1}{3}d \)
4. \( d = \frac{1}{3}e \)
5. \( e = \frac{1}{3}f \)

### Step 1: Express all variables in terms of \( f \)

Starting from \( e = \frac{1}{3}f \):
- \( d = \frac{1}{3}e = \frac{1}{3} \left(\frac{1}{3}f\right) = \frac{1}{9}f \)
- \( c = \frac{1}{3}d = \frac{1}{3} \left(\frac{1}{9}f\right) = \frac{1}{27}f \)
- \( b = \frac{1}{3}c = \frac{1}{3} \left(\frac{1}{27}f\right) = \frac{1}{81}f \)
- \( a = \frac{1}{3}b = \frac{1}{3} \left(\frac{1}{81}f\right) = \frac{1}{243}f \)

### Step 2: Substitute into the expression

Now we need to find the value of:
\[ \frac{a + b + c + d + e}{b + c + d + e + f} \]

Substituting the expressions we found:
\[ a + b + c + d + e = \frac{1}{243}f + \frac{1}{81}f + \frac{1}{27}f + \frac{1}{9}f + \frac{1}{3}f \]
\[ = \frac{1}{243}f + \frac{3}{243}f + \frac{9}{243}f + \frac{27}{243}f + \frac{81}{243}f \]
\[ = \frac{1 + 3 + 9 + 27 + 81}{243}f = \frac{121}{243}f \]

Now for the denominator:
\[ b + c + d + e + f = \frac{1}{81}f + \frac{1}{27}f + \frac{1}{9}f + \frac{1}{3}f + f \]
\[ = \frac{1}{81}f + \frac{3}{81}f + \frac{9}{81}f + \frac{27}{81}f + \frac{81}{81}f \]
\[ = \frac{1 + 3 + 9 + 27 + 81}{81}f = \frac{121}{81}f \]

### Step 3: Calculate the final expression

Now we can substitute these into our expression:
\[ \frac{a + b + c + d + e}{b + c + d + e + f} = \frac{\frac{121}{243}f}{\frac{121}{81}f} \]

The \( f \) cancels out:
\[ = \frac{121}{243} \times \frac{81}{121} = \frac{81}{243} = \frac{1}{3} \]

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