If in a G.P., a3 + a5 = 90 and if r = 2 find the first term of the G.P.

Answer

To solve the problem, we need to find the first term \( a \) of a geometric progression (G.P.) given that \( a_3 + a_5 = 90 \) and the common ratio \( r = 2 \).

### Step-by-Step Solution:

1. **Identify the terms in the G.P.**:
The \( n \)-th term of a G.P. can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
Therefore, we can express \( a_3 \) and \( a_5 \) as follows:
- \( a_3 = a \cdot r^{3-1} = a \cdot r^2 \)
- \( a_5 = a \cdot r^{5-1} = a \cdot r^4 \)

2. **Substituting the common ratio**:
Given \( r = 2 \), we substitute \( r \) into the expressions for \( a_3 \) and \( a_5 \):
- \( a_3 = a \cdot 2^2 = 4a \)
- \( a_5 = a \cdot 2^4 = 16a \)

3. **Set up the equation**:
According to the problem, we have:
\[ a_3 + a_5 = 90 \]
Substituting the expressions we found:
\[ 4a + 16a = 90 \]

4. **Combine like terms**:
Combine the terms on the left side:
\[ 20a = 90 \]

5. **Solve for \( a \)**:
To find \( a \), divide both sides by 20:
\[ a = \frac{90}{20} = \frac{9}{2} \]

### Final Answer:
The first term of the G.P. is:
\[ a = \frac{9}{2} \]