Question

Let \( a_1, a_2, a_3, \ldots \) be in an A.P. such that \[ \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1, \quad a_1 \neq 0. \] If \[ \sum_{k=1}^{n} a_k = 0, \] 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

We are given an Arithmetic Progression (A.P.) defined by \( a_1, a_2, a_3, \ldots \). Two conditions are provided: a sum involving specific terms of the A.P. and the sum of the first \( n \) terms being zero. We need to find the value of \( n \).

Concept Used:

The solution relies on the fundamental formulas for an Arithmetic Progression:

1. The k-th term of an A.P.: The general term \( a_k \) is given by \( a_k = a_1 + (k-1)d \), where \( a_1 \) is the first term and \( d \) is the common difference.

2. Sum of n terms of an A.P.: The sum of the first \( n \) terms, \( S_n \), is given by the formula:

\[ S_n = \frac{n}{2}[2a_1 + (n-1)d] \]

Alternatively, the sum of an A.P. with \( m \) terms can be expressed as \( S_m = \frac{m}{2}(\text{first term} + \text{last term}) \).

Step-by-Step Solution:

We start with the first given condition:

\[ \sum_{k=1}^{12} a_{2k - 1} = -\frac{72}{5} a_1 \]

The terms in this summation are \( a_1, a_3, a_5, \ldots, a_{23} \). This is a sequence of 12 terms which also forms an A.P. The first term of this new sequence is \( a_1 \), and the last term is \( a_{23} \). The common difference of this sequence is \( a_3 - a_1 = (a_1 + 2d) - a_1 = 2d \).

We can find the sum of these 12 terms using the formula \( S_m = \frac{m}{2}(\text{first term} + \text{last term}) \):

\[ \sum_{k=1}^{12} a_{2k - 1} = \frac{12}{2}(a_1 + a_{23}) = 6(a_1 + a_{23}) \]

We express \( a_{23} \) in terms of \( a_1 \) and \( d \): \( a_{23} = a_1 + (23-1)d = a_1 + 22d \).

\[ 6(a_1 + a_1 + 22d) = 6(2a_1 + 22d) = 12a_1 + 132d \]

Now, we equate this expression with the given condition:

\[ 12a_1 + 132d = -\frac{72}{5} a_1 \]

We solve this equation to find a relationship between \( d \) and \( a_1 \).

\[ 132d = -\frac{72}{5} a_1 - 12a_1 = \left(-\frac{72}{5} - \frac{60}{5}\right) a_1 = -\frac{132}{5} a_1 \]

Dividing both sides by 132, we get:

\[ d = -\frac{1}{5} a_1 \]

Next, we use the second condition:

\[ \sum_{k=1}^{n} a_k = 0 \]

Using the sum formula for the first \( n \) terms of the A.P.:

\[ S_n = \frac{n}{2}[2a_1 + (n-1)d] = 0 \]

Since \( n \) represents the number of terms, \( n \) cannot be 0. Thus, the term in the brackets must be zero:

\[ 2a_1 + (n-1)d = 0 \]

Final Computation & Result:

Now we substitute the relationship \( d = -\frac{1}{5} a_1 \) into the equation derived from the second condition:

\[ 2a_1 + (n-1)\left(-\frac{1}{5} a_1\right) = 0 \]

We are given that \( a_1 \neq 0 \), so we can divide the entire equation by \( a_1 \):

\[ 2 - \frac{n-1}{5} = 0 \]

Now, we solve for \( n \):

\[ 2 = \frac{n-1}{5} \] \[ 10 = n - 1 \] \[ n = 11 \]

Hence, the value of \( n \) is 11.

Answer 2

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: \[ \frac{n}{2}(2a + (n - 1)d) = 0 \Rightarrow (n - 11)d = 0 \Rightarrow n = 11 \]