The number of terms of an A.P. is even; the s | vedclass

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(A) Let the number of terms be $n = 2k$. The terms are $a_1, a_2, \ldots, a_{2k}$.Sum of even terms: $a_2 + a_4 + \ldots + a_{2k} = 30$.Sum of odd ter

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$4$

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(A) Let the number of terms be $n = 2k$. The terms are $a_1, a_2, \ldots, a_{2k}$.Sum of even terms: $a_2 + a_4 + \ldots + a_{2k} = 30$.Sum of odd terms: $a_1 + a_3 + \ldots + a_{2k-1} = 24$.Subtracting the two sums: $(a_2 - a_1) + (a_4 - a_3) + \ldots + (a_{2k} - a_{2k-1}) = 30 - 24 = 6$.Since each difference is the common difference $d$,we have $k 	imes d = 6$,so $n 	imes d = 2k 	imes d = 12$.The last term exceeds the first by $\frac{21}{2}$,so $a_n - a_1 = (n-1)d = \frac{21}{2}$.Substituting $nd = 12$: $12 - d = \frac{21}{2} \Rightarrow d = 12 - 10.5 = 1.5 = \frac{3}{2}$.Since $nd = 12$,$n 	imes \frac{3}{2} = 12 \Rightarrow n = 8$.Using the sum of odd terms: $\frac{k}{2}[2a_1 + (k-1)d] = 24$ where $k=4$ and $d=1.5$.$2[2a_1 + 3(1.5)] = 24$ $\Rightarrow 2a_1 + 4.5 = 12$ $\Rightarrow 2a_1 = 7.5$ $\Rightarrow a_1 = 3.75 = \frac{15}{4}$.Wait,re-evaluating $a_1$: $2[2a_1 + 4.5] = 24$ $\Rightarrow 2a_1 + 4.5 = 12$ $\Rightarrow 2a_1 = 7.5$ $\Rightarrow a_1 = 3.75$.Actually,using $a_1 + a_3 + a_5 + a_7 = 4a_1 + (0+2+4+6)d = 4a_1 + 12d = 24$.$4a_1 + 12(1.5) = 24$ $\Rightarrow 4a_1 + 18 = 24$ $\Rightarrow 4a_1 = 6$ $\Rightarrow a_1 = 1.5$.The sequence is $1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12$.The integer terms are $3, 6, 9, 12$. There are $4$ such terms.

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:

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