Suppose that the number of terms in an A.P. i | vedclass

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(A) Let the $A.P.$ be $a_1, a_2, a_3, \ldots, a_{2k}$.Given the sum of odd-positioned terms: $\sum_{r=1}^{k} a_{2r-1} = 40$.Given the sum of even-posi

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$5$

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(A) Let the $A.P.$ be $a_1, a_2, a_3, \ldots, a_{2k}$.Given the sum of odd-positioned terms: $\sum_{r=1}^{k} a_{2r-1} = 40$.Given the sum of even-positioned terms: $\sum_{r=1}^{k} a_{2r} = 55$.The difference between the sum of even and odd terms is $\sum_{r=1}^{k} (a_{2r} - a_{2r-1}) = 55 - 40 = 15$.Since $a_{2r} - a_{2r-1} = d$,we have $k 	imes d = 15$,so $d = \frac{15}{k}$.The last term is $a_{2k} = a_1 + (2k-1)d$. Given $a_{2k} - a_1 = 27$,we have $(2k-1)d = 27$.Substituting $d = \frac{15}{k}$ into the equation: $(2k-1) \frac{15}{k} = 27$.$15(2k-1) = 27k \Rightarrow 30k - 15 = 27k$.$3k = 15 \Rightarrow k = 5$.

Suppose that the number of terms in an $A.P.$ is $2k$,where $k \in N$. If the sum of all odd-positioned terms of the $A.P.$ is $40$,the sum of all even-positioned terms is $55$,and the last term of the $A.P.$ exceeds the first term by $27$,then $k$ is equal to:

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