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

Question

Accepted Answer

To determine how many terms of the arithmetic progression (A.P.) 2, 4, 6, 8, 10, ... are needed to achieve 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.

For this A.P., the first term $ a = 2 $ and the common difference $ d = 2 $.

We need to find $ n $ such that $ S_n = 210 $.

Substitute the known values into the formula:

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

Simplify the expression:

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

This simplifies to the quadratic equation:

$$ n^2 + n - 210 = 0 $$

To solve this quadratic equation, we can use the quadratic formula:

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

where $ a = 1 $, $ b = 1 $, and $ c = -210 $.

Calculate the discriminant:

$$ b^2 - 4ac = 1^2 - 4 \times 1 \times (-210) = 1 + 840 = 841 $$

Since the discriminant is a perfect square, we can find $ n $:

$$ n = \frac{-1 \pm \sqrt{841}}{2} $$
$$ n = \frac{-1 \pm 29}{2} $$

This gives two potential solutions for $ n $:

$$ n = \frac{28}{2} = 14 $$
$$ n = \frac{-30}{2} = -15 $$

Since $ n $ must be a positive integer, we have $ n = 14 $.

Therefore, 14 terms are needed to achieve a sum of 210, which corresponds to Option 4.

**Explanation for Other Options:**

1. **Option 1 (11 terms):** If we calculate the sum for 11 terms, $ S_{11} = \frac{11}{2} \times (2 \times 2 + (11-1) \times 2) = \frac{11}{2} \times (4 + 20) = \frac{11}{2} \times 24 = 132 $. This is less than 210.

2. **Option 2 (12 terms):** For 12 terms, $ S_{12} = \frac{12}{2} \times (2 \times 2 + (12-1) \times 2) = 6 \times (4 + 22) = 6 \times 26 = 156 $. This is also less than 210.

3. **Option 3 (13 terms):** For 13 terms, $ S_{13} = \frac{13}{2} \times (2 \times 2 + (13-1) \times 2) = \frac{13}{2} \times (4 + 24) = \frac{13}{2} \times 28 = 182 $. This is still less than 210.

Thus, only Option 4 (14 terms) gives the correct sum of 210, confirming it as the correct answer.