Question

Total number of terms in an A.P. are even. Sum of odd terms is 24 and sum of even terms is 30. Last term exceeds the first term by \( \frac{21}{2} \). Find the total number of terms.

• A. \( 10 \)
• B. \( 12 \)
• C. \( 14 \)
• D. \( 16 \)

Hint:

When dealing with sums of terms in an arithmetic progression, use the sum formula \( S_n = \frac{n}{2} \left( 2a + (n-1) d \right) \), and apply the conditions given in the problem to form equations.

Answer 1

The Correct Option is B

Let \( a \) be the first term of an arithmetic progression (A.P.) and \( d \) be the common difference. The total number of terms is even, denoted as \( 2n \). The sum of the odd-indexed terms is 24, and the sum of the even-indexed terms is 30. The difference between the last term and the first term is \( \frac{21}{2} \), leading to the equation \( a + (2n - 1) d = a + \frac{21}{2} \). Simplifying this yields \( (2n - 1) d = \frac{21}{2} \), and thus \( d = \frac{21}{2(2n - 1)} \). The sum of the first \( n \) odd terms is \( S_{\text{odd}} = \frac{n}{2} \left( 2a + (2n - 1)d \right) = 24 \). The sum of the first \( n \) even terms is \( S_{\text{even}} = \frac{n}{2} \left( 2a + 2nd \right) = 30 \). Solving these equations reveals that the total number of terms is \( 12 \).