Find the middle most term (s) of the $AP:-11,\ -7,\ -3,\ ..., 49$.

Given: An $AP:–11 –7,\ -3,\ ..., 49$.

To do: To find the middle most terms of the $AP$.

Solution:

First term $a=-11$, common difference $d=4$

$a_n=a+( n-1)d$

$\Rightarrow 49=-11+( n-1)4$

$\Rightarrow 60=4n-4$

$\Rightarrow 64=4n$

$\Rightarrow n=\frac{64}{4}$

$\Rightarrow n=16$  

Thus, there are $16$ terms.

Since the terms are even, there will be two middle most terms.

$\therefore n=8,\ 9$

When $n=8$

$a_8=-11+7\times4=17$

When $n=9$

$\Rightarrow a_9=-11+8\times4=21$

$\therefore a_{( 0,\ 9)}=\frac{17+21}{2}$

​

$=19$

Thus, the middle most term of the given A.P. is $19$.

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Updated on: 10-Oct-2022

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