The first term of a G.P. is 1. The sum of its third and fifth terms of 90. Find the common ratio of the G.P.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the first term and the terms of the G.P.
The first term of the G.P. is given as \( a = 1 \).

### Step 2: Write the expressions for the 3rd and 5th terms.
The \( n \)-th term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

For the 3rd term (\( T_3 \)):
\[ T_3 = a \cdot r^{3-1} = 1 \cdot r^2 = r^2 \]

For the 5th term (\( T_5 \)):
\[ T_5 = a \cdot r^{5-1} = 1 \cdot r^4 = r^4 \]

### Step 3: Set up the equation based on the sum of the 3rd and 5th terms.
According to the problem, the sum of the 3rd and 5th terms is 90:
\[ T_3 + T_5 = 90 \]
Substituting the expressions we found:
\[ r^2 + r^4 = 90 \]

### Step 4: Rearrange the equation.
Rearranging gives us:
\[ r^4 + r^2 - 90 = 0 \]

### Step 5: Substitute \( x = r^2 \).
Let \( x = r^2 \). Then the equation becomes:
\[ x^2 + x - 90 = 0 \]

### Step 6: Factor the quadratic equation.
To factor the quadratic equation, we need two numbers that multiply to \(-90\) and add to \(1\). The numbers \(10\) and \(-9\) work:
\[ (x + 10)(x - 9) = 0 \]

### Step 7: Solve for \( x \).
Setting each factor to zero gives us:
\[ x + 10 = 0 \quad \text{or} \quad x - 9 = 0 \]
This leads to:
\[ x = -10 \quad \text{or} \quad x = 9 \]

### Step 8: Substitute back to find \( r \).
Since \( x = r^2 \), we have:
1. \( r^2 = -10 \) (not valid since \( r^2 \) cannot be negative)
2. \( r^2 = 9 \)

Taking the square root of \( r^2 = 9 \):
\[ r = 3 \quad \text{or} \quad r = -3 \]

### Final Answer:
The common ratio \( r \) can be either \( 3 \) or \( -3 \).

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