If the first term in an arithmetic series is 3 and the last term is 136, and the sum is 1390, what are the first four terms ?

Options

A. 3, 9, 27, 81
B. 3, 10, 17, 24
C. 3, 12, 36, 94
D. 3, 14 , 28, 108

Answer

To find the first four terms of the arithmetic series given the first term, last term, and the sum, we can follow these steps:

### Step 1: Identify the known values
- First term (a) = 3
- Last term (L) = 136
- Sum of the series (S_n) = 1390

### Step 2: Use the formula for the sum of an arithmetic series
The formula for the sum of the first n terms of an arithmetic series is:

\[ S_n = \frac{n}{2} (a + L) \]

Substituting the known values into the formula:

\[ 1390 = \frac{n}{2} (3 + 136) \]

### Step 3: Simplify the equation
Calculate \(3 + 136\):

\[ 3 + 136 = 139 \]

Now substitute this back into the equation:

\[ 1390 = \frac{n}{2} \times 139 \]

### Step 4: Solve for n
Multiply both sides by 2 to eliminate the fraction:

\[ 2780 = n \times 139 \]

Now, divide both sides by 139 to find n:

\[ n = \frac{2780}{139} = 20 \]

### Step 5: Use the last term formula to find the common difference (d)
The last term of an arithmetic series can be expressed as:

\[ L = a + (n - 1)d \]

Substituting the known values:

\[ 136 = 3 + (20 - 1)d \]

This simplifies to:

\[ 136 = 3 + 19d \]

### Step 6: Solve for d
Subtract 3 from both sides:

\[ 133 = 19d \]

Now divide by 19:

\[ d = \frac{133}{19} = 7 \]

### Step 7: Find the first four terms
Now that we have \(a = 3\) and \(d = 7\), we can find the first four terms of the series:

1. First term (a) = 3
2. Second term = \(a + d = 3 + 7 = 10\)
3. Third term = \(a + 2d = 3 + 2 \times 7 = 3 + 14 = 17\)
4. Fourth term = \(a + 3d = 3 + 3 \times 7 = 3 + 21 = 24\)

### Final Answer
The first four terms of the arithmetic series are: **3, 10, 17, 24**

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