Find the common ratio of the G.P. whose first and last terms are 5 and `(32)/(625)` respectively and the sum of the G.P. is `(5187)/(625)`:

Options

A. `1/5`
B. `2/5`
C. `5/3`
D. `4/5`

Answer

To find the common ratio of the geometric progression (G.P.) whose first term \( a = 5 \), last term \( a_n = \frac{32}{625} \), and sum \( S_n = \frac{5187}{625} \), we can follow these steps:

### Step 1: Write down the formulas
The last term of a G.P. can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
where \( r \) is the common ratio and \( n \) is the number of terms.

The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
provided \( r \neq 1 \).

### Step 2: Substitute known values into the last term formula
From the last term formula, we have:
\[ \frac{32}{625} = 5 \cdot r^{n-1} \]
To isolate \( r^{n-1} \), divide both sides by 5:
\[ r^{n-1} = \frac{32}{625} \cdot \frac{1}{5} = \frac{32}{3125} \]

### Step 3: Substitute known values into the sum formula
Now, substituting into the sum formula:
\[ \frac{5187}{625} = \frac{5(1 - r^n)}{1 - r} \]
Multiply both sides by \( 625(1 - r) \):
\[ 5187(1 - r) = 5(1 - r^n) \cdot 625 \]
This simplifies to:
\[ 5187 - 5187r = 3125 - 3125r^n \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 5187 - 3125 = 5187r - 3125r^n \]
Thus:
\[ 2062 = 5187r - 3125r^n \]

### Step 5: Substitute \( r^n \) in terms of \( r \)
From the earlier step, we know \( r^{n-1} = \frac{32}{3125} \). Therefore:
\[ r^n = r \cdot r^{n-1} = r \cdot \frac{32}{3125} \]
Substituting this into the equation gives:
\[ 2062 = 5187r - 3125 \cdot r \cdot \frac{32}{3125} \]
This simplifies to:
\[ 2062 = 5187r - 32r \]
\[ 2062 = (5187 - 32)r \]
\[ 2062 = 5155r \]

### Step 6: Solve for \( r \)
Now, divide both sides by 5155:
\[ r = \frac{2062}{5155} \]
This can be simplified:
\[ r = \frac{1031 \cdot 2}{1031 \cdot 5} = \frac{2}{5} \]

### Final Answer
Thus, the common ratio \( r \) of the G.P. is:
\[ \boxed{\frac{2}{5}} \]