Let `a_(1),a_(2),a_(3)"……."` be an arithmetic progression and `b_(1), b_(2), b_(3), "……."` be a geometric progression 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 value of sum of `sum_(i = 1)^(20) a_(i)` is equal to
`"(a) 480 (b) 770 (c) 960 (d) 1040"`

Options

A. 480
B. 770
C. 960
D. 1040

Answer

To solve the problem step by step, we will follow the given information about the arithmetic progression (AP) and geometric progression (GP) to find the required sum.

### Step 1: Define the sequences
Let:
- \( a_1, a_2, a_3, \ldots \) be an arithmetic progression with first term \( a \) and common difference \( d \).
- Thus, we have:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- \( a_4 = a + 3d \)

- \( b_1, b_2, b_3, \ldots \) be a geometric progression with first term \( A \) and common ratio \( r \).
- Thus, we have:
- \( b_1 = A \)
- \( b_2 = Ar \)
- \( b_3 = Ar^2 \)
- \( b_4 = Ar^3 \)

### Step 2: Write the equations for \( c_n \)
The sequence \( c_n \) is defined as:
\[ c_n = a_n + b_n \]
From the problem, we have:
- \( c_1 = a + A = 1 \) (Equation 1)
- \( c_2 = (a + d) + Ar = 4 \) (Equation 2)
- \( c_3 = (a + 2d) + Ar^2 = 15 \) (Equation 3)
- \( c_4 = (a + 3d) + Ar^3 = 2 \) (Equation 4)

### Step 3: Set up the equations
From the equations we have:
1. \( a + A = 1 \) (1)
2. \( a + d + Ar = 4 \) (2)
3. \( a + 2d + Ar^2 = 15 \) (3)
4. \( a + 3d + Ar^3 = 2 \) (4)

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

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

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

### Step 5: Solve for \( A \) and \( r \)
From Equation 1:
\[ A = 1 - a \]

Substituting \( A \) into Equation 7:
\[ d + Ar - (1 - a) = 3 \]
This gives:
\[ d + Ar + a - 1 = 3 \implies d + Ar + a = 4 \quad \text{(Equation 8)} \]

### Step 6: Substitute and solve for \( a \), \( d \), and \( A \)
Now we have three equations (5, 6, and 8) with three unknowns \( a, d, A \).

1. From Equation 5:
\[ d + Ar^2 - Ar = 11 \]
2. From Equation 6:
\[ d + Ar^3 - Ar^2 = -13 \]

We can solve these equations simultaneously to find \( a \), \( d \), and \( A \).

### Step 7: Calculate the sum \( S_{20} \)
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \left(2a + (20-1)d\right) = 10 \left(2a + 19d\right) \]

### Step 8: Substitute the values of \( a \) and \( d \)
After solving the equations, we find:
- \( a = \frac{1}{2} \)
- \( d = 5 \)

Substituting these values into the sum formula:
\[ S_{20} = 10 \left(2 \cdot \frac{1}{2} + 19 \cdot 5\right) = 10 \left(1 + 95\right) = 10 \cdot 96 = 960 \]

### Final Answer
The value of the sum \( \sum_{i=1}^{20} a_i \) is **960**.