Question

The value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is ________.

Answer 1

The given A.P. is: –11, –7, –3, ...., 49, 53

First term (a) = –11
Common difference (d) = a₂ – a₁
= –7 – (–11)
= –7 + 11
= 4
Last term (a_n) = 53

Now,
$$\begin{gathered} a_n=a+\left(n-1\right)d \\ \Rightarrow 53=-11+\left(n-1\right)\left(4\right) \\ \Rightarrow 53=-11+4n-4 \\ \Rightarrow 53=-15+4n \\ \Rightarrow 4n=53+15 \\ \Rightarrow 4n=68 \\ \Rightarrow n=17 \end{gathered}$$

∴ Total number of terms = 17

$$\begin{gathered} \mathrm{Middle}\mathrm{term}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}}\mathrm{term} \\ ={\left(\frac{17+!}{2}\right)}^{\mathrm{th}}\mathrm{term} \\ =9^{\mathrm{th}}\mathrm{term} \end{gathered}$$

Therefore,
$$\begin{gathered} a_9=a+\left(9-1\right)d \\ =-11+8\left(4\right) \\ =-11+32 \\ =21 \end{gathered}$$

​Hence, the value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is 21.