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 frac212$\frac{21}{2}$. Then the number of terms which are integers in the A.P. is :

### Related Formula textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right)$$\text{Sum of an A.P.: } S_k = \frac{k}{2} \left( 2a + (k-1)d \right)$$ textGeneral term of an A.P.: a_k = a_1 + (k-1)d$$\text{General term of an A.P.: } a_k = a_1 + (k-1)d$$ ### Core Logic Let the A.P. have n$n$ terms (where n$n$ is even). The terms are divided into n/2$n/2$ odd-indexed terms and n/2$n/2$ even-indexed terms. ### Step 1: Set up the even and odd sums Sum of even terms: a_2 + a_4 + dots + a_n = 30 quad text--- (1)$$a_2 + a_4 + \dots + a_n = 30 \quad \text{--- (1)}$$ Sum of odd terms: a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2)$$a_1 + a_3 + \dots + a_{n-1} = 24 \quad \text{--- (2)}$$ Subtracting equation (2) from (1): (a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6$$(a_2 - a_1) + (a_4 - a_3) + \dots + (a_n - a_{n-1}) = 30 - 24 = 6$$ Since there are n/2$n/2$ such pairs, and the difference of adjacent terms is the common difference d$d$: fracn2 d = 6 implies n d = 12 quad text--- (3)$$\frac{n}{2} d = 6 \implies n d = 12 \quad \text{--- (3)}$$ ### Step 2: Solve for n and d We are given that the last term exceeds the first by frac212$\frac{21}{2}$: a_n - a_1 = (n-1)d = frac212$$a_n - a_1 = (n-1)d = \frac{21}{2}$$ n d - d = 10.5$n d - d = 10.5$ Substitute nd = 12$nd = 12$ from (3): 12 - d = 10.5 implies d = 1.5 = frac32$$12 - d = 10.5 \implies d = 1.5 = \frac{3}{2}$$ Using this in (3): n left(frac32right) = 12 implies n = 8$$n \left(\frac{3}{2}\right) = 12 \implies n = 8$$ ### Step 3: Solve for the first term The sum of the odd terms is: S_textodd = frac42 left[ 2a_1 + (4-1)(2d) right] = 24$$S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1)(2d) \right] = 24$$ 2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32$$2 \left[ 2a_1 + 3(3) \right] = 24 \implies 2a_1 + 9 = 12 \implies a_1 = 1.5 = \frac{3}{2}$$ Thus, the terms are: frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12$$\frac{3}{2}, \, 3, \, \frac{9}{2}, \, 6, \, \frac{15}{2}, \, 9, \, \frac{21}{2}, \, 12$$ The terms that are integers are 3, 6, 9, 12$3, 6, 9, 12$. The total number of integer terms is 4. ### Pattern Recognition Sum of even terms minus sum of odd terms in any A.P. with an even number of terms n$n$ is always equal to fracn2 d$\frac{n}{2} d$. This is an extremely useful relation to remember. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series