If `b_(1), b_(2), b_(3),….` belongs to an A.P. such that `b_(1)+b_(4)+b_(7)+…+b_(28)=220`, then the value of `b_(1)+b_(2)+b_(3)+…+b_(28)=`

Options

A. 616
B. 308
C. 2200
D. 1232

Answer

To solve the problem, we need to find the sum of the first 28 terms of an arithmetic progression (AP) given that the sum of every third term starting from the first term equals 220. Let's break it down step by step.

### Step 1: Understand the terms of the AP
Let the first term of the AP be \( b_1 = a \) and the common difference be \( d \). The terms of the AP can be expressed as:
- \( b_1 = a \)
- \( b_2 = a + d \)
- \( b_3 = a + 2d \)
- \( b_4 = a + 3d \)
- \( b_5 = a + 4d \)
- ...
- \( b_n = a + (n-1)d \)

### Step 2: Identify the terms to sum
We need to find the sum of the terms \( b_1, b_4, b_7, \ldots, b_{28} \). These terms can be expressed as:
- \( b_1 = a \)
- \( b_4 = a + 3d \)
- \( b_7 = a + 6d \)
- ...
- \( b_{28} = a + 27d \)

### Step 3: Determine the number of terms
The sequence of indices for the terms we are summing is \( 1, 4, 7, \ldots, 28 \). This is an arithmetic sequence where:
- First term \( n_1 = 1 \)
- Common difference \( = 3 \)
- Last term \( n_k = 28 \)

To find the number of terms \( n \):
\[ n = \frac{28 - 1}{3} + 1 = \frac{27}{3} + 1 = 9 + 1 = 10 \]

### Step 4: Write the sum of the selected terms
The sum of these 10 terms can be expressed as:
\[ S = b_1 + b_4 + b_7 + \ldots + b_{28} = 10a + (3d + 6d + 9d + \ldots + 27d) \]
The second part can be simplified:
\[ 3d(1 + 2 + 3 + \ldots + 9) = 3d \cdot \frac{9 \cdot 10}{2} = 3d \cdot 45 = 135d \]
Thus, we have:
\[ S = 10a + 135d = 220 \quad (1) \]

### Step 5: Find the sum of the first 28 terms
Now, we need to find the sum of the first 28 terms of the AP:
\[ S_{28} = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{28}{2} \cdot (2a + 27d) = 14(2a + 27d) \]

### Step 6: Relate \( 2a + 27d \) to the previous equation
From equation (1), we can rearrange it:
\[ 10a + 135d = 220 \]
To express \( 2a + 27d \), we can manipulate the equation:
\[ 2a + 27d = \frac{2}{10}(10a) + \frac{27}{135}(135d) = \frac{2}{10}(220 - 135d) + \frac{27}{135}(135d) \]
However, it is simpler to directly substitute values. We can express \( 2a + 27d \) in terms of \( 10a + 135d \).

### Step 7: Solve for \( S_{28} \)
Using the values we derived:
\[ 2a + 27d = \frac{220 - 135d}{10} \cdot 2 + 27d \]
We can simplify this further or directly substitute \( 2a + 27d \) into \( S_{28} \):
\[ S_{28} = 14(2a + 27d) = 14 \cdot 44 = 616 \]

### Final Answer
Thus, the sum of the first 28 terms of the AP is:
\[ \boxed{616} \]