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8. Sequences and Series
medium
Suppose that the number of terms in an $A.P.$ is $2 k$, $k \in N$. If the sum of all odd terms of the $A.P.$ is $40 ,$ the sum of all even terms is $55$ and the last term of the $A.P.$ exceeds the first term by $27$ , then $k$ is equal to
A$5$
B$8$
C$6$
D$4$
(JEE MAIN-2025)
Solution
$ a _1, a _2, a _3, \ldots \ldots, a _{2 k }$
$\sum_{ r =1}^{ k } a _{2 r -1}=40, \sum_{ r =1}^{ k } a _{2 r }=55, a _{2 k }- a _1=27$
$\frac{ k }{2}\left[2 a _1+( k -1) 2 d\right]=40, \frac{ k }{2}\left[2 a _2+( k -1) 2 d\right]=55,$
$d=\frac{27}{2 k -1}$
$a _1=\frac{40}{ k }-( k -1) d =\frac{55}{ k }- kd$
$d =\frac{15}{ k } \Rightarrow \frac{27}{2 k -1}=\frac{15}{ k } \Rightarrow 9 k =10 k -5$
$\therefore k =5$
$\sum_{ r =1}^{ k } a _{2 r -1}=40, \sum_{ r =1}^{ k } a _{2 r }=55, a _{2 k }- a _1=27$
$\frac{ k }{2}\left[2 a _1+( k -1) 2 d\right]=40, \frac{ k }{2}\left[2 a _2+( k -1) 2 d\right]=55,$
$d=\frac{27}{2 k -1}$
$a _1=\frac{40}{ k }-( k -1) d =\frac{55}{ k }- kd$
$d =\frac{15}{ k } \Rightarrow \frac{27}{2 k -1}=\frac{15}{ k } \Rightarrow 9 k =10 k -5$
$\therefore k =5$
Standard 11
Mathematics
