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. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Answer 2

Let the number of terms in the A.P. be $2n$. Let the first term be $a$ and the common difference be $d$. The sum of the odd-indexed terms is $n[a + (n-1)d] = 24$ --- (1) The sum of the even-indexed terms is $n[a+nd] = 30$ --- (2) The last term exceeds the first term by $(2n-1)d = \frac{21}{2}$ --- (3)

Subtracting equation (1) from equation (2) gives $nd = 6$ --- (4)

Dividing equation (3) by equation (4): $\frac{(2n-1)d}{nd} = \frac{21/2}{6} \implies \frac{2n-1}{n} = \frac{7}{4}$ $8n - 4 = 7n \implies n = 4$.

The total number of terms is $2n = 8$. From $nd = 6$, we get $4d = 6 \implies d = \frac{3}{2}$. From equation (1), $4[a + (4-1)\frac{3}{2}] = 24 \implies 4[a + \frac{9}{2}] = 24 \implies a + \frac{9}{2} = 6 \implies a = \frac{3}{2}$.

The terms of the A.P. are $a_k = a + (k-1)d = \frac{3}{2} + (k-1)\frac{3}{2} = \frac{3k}{2}$. For $a_k$ to be an integer, $k$ must be even. For $k \in \{1, 2, \dots, 8\}$, the even values of $k$ are $\{2, 4, 6, 8\}$. Thus, there are 4 terms that are integers.