Question

Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:

• A. 11
• B. 10
• C. 18
• D. 17

Hint:

For A.P. sums, use known identities and match values step-by-step. Cross-check with total sum condition.

Answer 1

The Correct Option is A

Let first term be \( a \), common difference = \( d \). Sum of 12 odd-position terms: \[ 2 \cdot \left[ \frac{12}{2} \cdot \left( 2a + (12 - 1)d \right) \right] = \frac{72}{5} \Rightarrow 12a + 132d = \frac{72}{5} \Rightarrow 60a + 660d = 72 \Rightarrow a = -5d \] Total sum up to \( n \) terms: \[ \frac{n}{2}(2a + (n - 1)d) = 0 \Rightarrow (n - 11)d = 0 \Rightarrow n = 11 \]