The sum of the first ten terms of an A.P. , equals 155
and the sum of the first two terms of a G.P. equals
9. The first term of the A.P. is equal to the common
ratio of the G.P. and the common difference of the
A.P. is equal to the first term G.P.. Give that the
common difference of the A.P. is less then unity,
which of the following is correct ?

Options

A. `(25)/(2)` is a term of the G.P.
B. `(28)/(3)` is a term of the G.P.
C. `(28)/(3)` is a term of the A.P.
D. `(25)/(2)` is a term of the A.P.

Answer

To solve the problem step by step, we will break down the information given and apply the formulas for the sum of an arithmetic progression (A.P.) and a geometric progression (G.P.).

### Step 1: Understand the given information
We have:
1. The sum of the first 10 terms of an A.P. is 155.
2. The sum of the first two terms of a G.P. is 9.
3. The first term of the A.P. (let's denote it as \( A \)) is equal to the common ratio of the G.P. (denote it as \( R \)).
4. The common difference of the A.P. (denote it as \( D \)) is equal to the first term of the G.P. (denote it as \( B \)).
5. The common difference \( D < 1 \).

### Step 2: Write the equations based on the information
1. The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]
For \( n = 10 \):
\[ S_{10} = \frac{10}{2} \times (2A + 9D) = 155 \]
Simplifying this:
\[ 5(2A + 9D) = 155 \implies 2A + 9D = 31 \quad \text{(Equation 1)} \]

2. The formula for the sum of the first \( n \) terms of a G.P. is:
\[ S_n = A \frac{(r^n - 1)}{r - 1} \]
For \( n = 2 \):
\[ S_2 = B(1 + R) = 9 \quad \text{(Equation 2)} \]

### Step 3: Substitute the relationships
From the problem statement:
- \( A = R \)
- \( D = B \)

Substituting \( R \) and \( B \) into Equation 2:
\[ B(1 + A) = 9 \implies D(1 + A) = 9 \quad \text{(Equation 3)} \]

### Step 4: Substitute \( D \) from Equation 3 into Equation 1
From Equation 3, we can express \( D \):
\[ D = \frac{9}{1 + A} \]

Now substitute \( D \) into Equation 1:
\[ 2A + 9\left(\frac{9}{1 + A}\right) = 31 \]
Multiply through by \( 1 + A \) to eliminate the fraction:
\[ 2A(1 + A) + 81 = 31(1 + A) \]
Expanding gives:
\[ 2A + 2A^2 + 81 = 31 + 31A \]
Rearranging:
\[ 2A^2 - 29A + 50 = 0 \quad \text{(Equation 4)} \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -29, c = 50 \):
\[ A = \frac{29 \pm \sqrt{(-29)^2 - 4 \cdot 2 \cdot 50}}{2 \cdot 2} \]
Calculating the discriminant:
\[ A = \frac{29 \pm \sqrt{841 - 400}}{4} = \frac{29 \pm \sqrt{441}}{4} = \frac{29 \pm 21}{4} \]
This gives us two solutions:
1. \( A = \frac{50}{4} = 12.5 \)
2. \( A = \frac{8}{4} = 2 \)

### Step 6: Find corresponding values of \( D \)
Using \( D = \frac{9}{1 + A} \):
1. If \( A = 12.5 \):
\[ D = \frac{9}{1 + 12.5} = \frac{9}{13.5} = \frac{2}{3} \]
2. If \( A = 2 \):
\[ D = \frac{9}{1 + 2} = \frac{9}{3} = 3 \]

### Step 7: Check the condition \( D < 1 \)
Since \( D = 3 \) is not less than 1, we discard it. Thus, we have:
- \( A = 12.5 \) and \( D = \frac{2}{3} \).

### Conclusion
The common difference of the A.P. is \( \frac{2}{3} \) and the first term of the A.P. is \( 12.5 \).