The sum of first ten terms of an A.P. is 155 and the sum of first two terms of a G.P. 9. The first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P. which can be the A.P. as per the given conditions?

Options

A. 2,4,6,8,10,.....
B. 25/2,79/6,83/6, .....
C. 2,5,8,11,.....
D. both (b) and (c)

Answer

To solve the problem, we need to find the possible arithmetic progression (A.P.) that satisfies the given conditions. Let's break it down step by step.

### Step 1: Define Variables
Let:
- \( a \) = first term of the A.P.
- \( d \) = common difference of the A.P.
- \( r \) = common ratio of the G.P.
- \( b \) = first term of the G.P.

From the problem, we know:
- \( a = r \) (the first term of the A.P. is equal to the common ratio of the G.P.)
- \( b = d \) (the first term of the G.P. is equal to the common difference of the A.P.)

### Step 2: Use the Sum of A.P.
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
For the first 10 terms of the A.P., we have:
\[ S_{10} = \frac{10}{2} \left(2a + 9d\right) = 5(2a + 9d) \]
We know from the problem that \( S_{10} = 155 \):
\[ 5(2a + 9d) = 155 \]
Dividing both sides by 5:
\[ 2a + 9d = 31 \quad \text{(Equation 1)} \]

### Step 3: Use the Sum of G.P.
The sum of the first two terms of a G.P. is:
\[ S_2 = b + br = b(1 + r) \]
We know \( S_2 = 9 \):
\[ b(1 + r) = 9 \]
Substituting \( b = d \) and \( r = a \):
\[ d(1 + a) = 9 \quad \text{(Equation 2)} \]

### Step 4: Substitute and Solve
From Equation 1:
\[ 2a + 9d = 31 \quad \text{(1)} \]
From Equation 2:
\[ d(1 + a) = 9 \quad \text{(2)} \]
We can express \( d \) from Equation 2:
\[ d = \frac{9}{1 + a} \]
Now, substitute this expression for \( d \) 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 both sides:
\[ 2a + 2a^2 + 81 = 31 + 31a \]
Rearranging gives:
\[ 2a^2 - 29a + 50 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 2, B = -29, C = 50 \):
\[ D = B^2 - 4AC = (-29)^2 - 4 \cdot 2 \cdot 50 = 841 - 400 = 441 \]
\[ a = \frac{29 \pm \sqrt{441}}{4} = \frac{29 \pm 21}{4} \]
Calculating the two possible values:
1. \( a = \frac{50}{4} = 12.5 \)
2. \( a = \frac{8}{4} = 2 \)

### Step 6: Find Corresponding Values of \( d \)
For \( a = 2 \):
\[ d = \frac{9}{1 + 2} = 3 \]
So, the A.P. is:
\[ 2, 5, 8, 11, \ldots \]

For \( a = 12.5 \):
\[ d = \frac{9}{1 + 12.5} = \frac{9}{13.5} = \frac{2}{3} \]
So, the A.P. is:
\[ 12.5, 13.5, 14.5, \ldots \]

### Conclusion
Thus, the possible A.P.s that satisfy the given conditions are:
1. \( 2, 5, 8, 11, \ldots \)
2. \( 12.5, 13.5, 14.5, \ldots \)