How many the first terms of the A.P. 2, 4, 6, 8, 10, _____ are needed to get sum 210?

Options

A. 11
B. 12
C. 13
D. 14

Answer

To find how many terms of the arithmetic progression (A.P.) 2, 4, 6, 8, 10, ... are needed to get a sum of 210, we can use the formula for the sum of the first n terms of an A.P.:

\[ S_n = \frac{n}{2} \times (2A + (n-1)d) \]

Where:
- \( S_n \) is the sum of the first n terms,
- \( A \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

### Step 1: Identify the values
From the given A.P., we have:
- First term \( A = 2 \)
- Common difference \( d = 4 - 2 = 2 \)

### Step 2: Set up the equation
We know the sum \( S_n = 210 \). Plugging in the values into the sum formula, we get:

\[ 210 = \frac{n}{2} \times (2 \times 2 + (n-1) \times 2) \]

### Step 3: Simplify the equation
Now, simplify the equation:

\[ 210 = \frac{n}{2} \times (4 + 2(n-1)) \]
\[ 210 = \frac{n}{2} \times (4 + 2n - 2) \]
\[ 210 = \frac{n}{2} \times (2n + 2) \]
\[ 210 = \frac{n}{2} \times 2(n + 1) \]
\[ 210 = n(n + 1) \]

### Step 4: Rearrange the equation
Rearranging gives us:

\[ n^2 + n - 210 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 1, b = 1, c = -210 \):

\[ n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-210)}}{2 \cdot 1} \]
\[ n = \frac{-1 \pm \sqrt{1 + 840}}{2} \]
\[ n = \frac{-1 \pm \sqrt{841}}{2} \]
\[ n = \frac{-1 \pm 29}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{28}{2} = 14 \)
2. \( n = \frac{-30}{2} = -15 \) (not valid since n must be positive)

### Step 6: Conclusion
Thus, the number of terms needed to get a sum of 210 is:

\[ n = 14 \]