Question

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $ \frac{21}{2} $. Then the number of terms which are integers in the A.P. is:

• A. 4
• B. 10
• C. 6
• D. 8

Hint:

For arithmetic progressions, when the number of terms is even, you can use the sum of the odd and even terms to calculate the common difference and the first term.

Answer 1

The Correct Option is A

The sum of the even terms: \[ a_2 + a_4 + \cdots + a_{2n} = 30 \quad \text{(Equation 1)} \] The sum of the odd terms: \[ a_1 + a_3 + \cdots + a_{2n-1} = 24 \quad \text{(Equation 2)} \] Subtracting Equation 2 from Equation 1: \[ (a_2 - a_1) + (a_4 - a_3) + \cdots + (a_{2n} - a_{2n-1}) = 6 \] \[ \frac{n}{2} \cdot d = 6 \quad \text{(where \( d \) is the common difference)} \] \[ n d = 6 \quad \Rightarrow \quad n = 12 \] From the equation \( a_{n} - a_1 = \frac{21}{2} \): \[ nd - d = \frac{21}{2} \quad \Rightarrow \quad 12d - d = \frac{21}{2} \] \[ 11d = \frac{21}{2} \quad \Rightarrow \quad d = \frac{3}{2} \] The sum of odd terms: \[ S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1) \cdot d \right] = 24 \] \[ a_1 = \frac{3}{2} \] The A.P. is: \( \frac{3}{2}, \, 3, \, 9, \, 12, 15, 21, 9, 21, \dots \) Thus, the number of terms is 4.