Deriving Number of Terms
The difference between the sum of even terms and the sum of odd terms can be expressed as: $S_{even} - S_{odd} = (a_2 - a_1) + (a_4 - a_3) + ... + (a_{2n} - a_{2n-1})$ Each pair $(a_{2k} - a_{2k-1})$ is equal to the common difference $d$. Since there are $n$ such pairs, we have: $S_{even} - S_{odd} = n \times d$ $30 - 24 = nd$ $6 = nd \quad (*)$ Also, the definition of the last term gives: $a_{2n} = a_1 + (2n-1)d$ $a_{2n} - a_1 = (2n-1)d$ $\frac{21}{2} = (2n-1)d \quad (**)$ Now we have a system of two equations with two unknowns ($n$ and $d$): 1. $nd = 6 \implies d = \frac{6}{n}$ 2. $(2n-1)d = \frac{21}{2}$ Substitute $d$ from the first equation into the second: $(2n-1) \left(\frac{6}{n}\right) = \frac{21}{2}$ $\frac{12n - 6}{n} = \frac{21}{2}$ $2(12n - 6) = 21n$ $24n - 12 = 21n$ $3n = 12$ $n = 4$ The total number of terms in the A.P. is $2n = 2 \times 4 = 8$.Calculating First Term and Common Difference
Using $n=4$ in equation $(*)$: $4d = 6$ $d = \frac{6}{4} = \frac{3}{2}$ The sum of all terms is $S_{2n} = S_{odd} + S_{even} = 24 + 30 = 54$. The sum formula is $S_{2n} = \frac{2n}{2}(a_1 + a_{2n}) = n(a_1 + a_{2n})$. $54 = 4(a_1 + a_{2n})$ $a_1 + a_{2n} = \frac{54}{4} = \frac{27}{2}$ We have: $a_1 + a_{2n} = \frac{27}{2}$ $a_{2n} - a_1 = \frac{21}{2}$ Adding these two equations: $2a_{2n} = \frac{27}{2} + \frac{21}{2} = \frac{48}{2} = 24 \implies a_{2n} = 12$. Subtracting the second from the first: $2a_1 = \frac{27}{2} - \frac{21}{2} = \frac{6}{2} = 3 \implies a_1 = \frac{3}{2}$.Identifying Integer Terms
The A.P. has $a_1 = \frac{3}{2}$ and $d = \frac{3}{2}$. The number of terms is 8. The terms are: $a_1 = \frac{3}{2}$ $a_2 = \frac{3}{2} + \frac{3}{2} = \frac{6}{2} = 3$ $a_3 = 3 + \frac{3}{2} = \frac{9}{2}$ $a_4 = \frac{9}{2} + \frac{3}{2} = \frac{12}{2} = 6$ $a_5 = 6 + \frac{3}{2} = \frac{15}{2}$ $a_6 = \frac{15}{2} + \frac{3}{2} = \frac{18}{2} = 9$ $a_7 = 9 + \frac{3}{2} = \frac{21}{2}$ $a_8 = \frac{21}{2} + \frac{3}{2} = \frac{24}{2} = 12$ The terms of the A.P. are: $\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$. The integer terms are $3, 6, 9, 12$. There are 4 integer terms.Similar Questions
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Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
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