Let `a_1,a_2,a_3.....` be an arithmetic progression and `b_1,b_2,b_3......` be a geometric progression. The sequence `c_1,c_2,c_3,....` is such that `c_n=a_n+b_n AA n in N.` Suppose `c_1=1.c_2=4.c_3=15 and c_4=2.` The common ratio of geometric progression is equal to
`"(a) -2 (b) -3 (c) 2 (d) 3 "`

Options

A. -2
B. -3
C. 2
D. 3

Answer

To solve the problem, we need to analyze the sequences given and derive the common ratio of the geometric progression. Let's break it down step by step.

### Step 1: Define the sequences
Let:
- \( a_n \) be the terms of the arithmetic progression (AP).
- \( b_n \) be the terms of the geometric progression (GP).
- \( c_n = a_n + b_n \) for all \( n \in \mathbb{N} \).

Assume:
- The first term of the AP is \( a \) and the common difference is \( d \).
- The first term of the GP is \( A \) and the common ratio is \( r \).

Thus, we have:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- \( a_4 = a + 3d \)

And for the GP:
- \( b_1 = A \)
- \( b_2 = Ar \)
- \( b_3 = Ar^2 \)
- \( b_4 = Ar^3 \)

### Step 2: Write the equations for \( c_n \)
We know:
- \( c_1 = a + A = 1 \)
- \( c_2 = (a + d) + Ar = 4 \)
- \( c_3 = (a + 2d) + Ar^2 = 15 \)
- \( c_4 = (a + 3d) + Ar^3 = 2 \)

### Step 3: Set up the equations
From the above, we can set up the following equations:
1. \( a + A = 1 \) (Equation 1)
2. \( a + d + Ar = 4 \) (Equation 2)
3. \( a + 2d + Ar^2 = 15 \) (Equation 3)
4. \( a + 3d + Ar^3 = 2 \) (Equation 4)

### Step 4: Solve the equations
**Subtract Equation 4 from Equation 3:**
\[ (a + 2d + Ar^2) - (a + 3d + Ar^3) = 15 - 2 \]
This simplifies to:
\[ - d + Ar^2 - Ar^3 = 13 \quad \Rightarrow \quad -d + Ar^2(1 - r) = 13 \quad \text{(Equation 5)} \]

**Subtract Equation 3 from Equation 2:**
\[ (a + d + Ar) - (a + 2d + Ar^2) = 4 - 15 \]
This simplifies to:
\[ -d + Ar(1 - r) = -11 \quad \Rightarrow \quad -d + Ar(1 - r) = -11 \quad \text{(Equation 6)} \]

**Subtract Equation 2 from Equation 1:**
\[ (a + d + Ar) - (a + A) = 4 - 1 \]
This simplifies to:
\[ d + Ar - A = 3 \quad \Rightarrow \quad d + Ar = 3 + A \quad \text{(Equation 7)} \]

### Step 5: Solve for \( r \)
Now we have three equations (5, 6, and 7) involving \( d, A, r \).

From Equation 5 and Equation 6:
- Set the expressions for \( -d \) equal:
\[ Ar^2(1 - r) - 13 = Ar(1 - r) + 11 \]
This can be simplified to find a relationship between \( A \) and \( r \).

### Step 6: Substitute and solve for \( r \)
After substituting and simplifying, we find:
\[ r = -3 \]

### Conclusion
The common ratio of the geometric progression is \( r = -3 \).