Question

Find the middle term of the AP 10, 7, 4, ..., (−62). [CBSE 2009C]

Answer 1

The given AP is 10, 7, 4, ..., −62.

First term, a = 10

Common difference, d = 7 − 10 = −3

Suppose there are n terms in the given AP. Then,

$$\begin{gathered} a_n=-62 \\ \Rightarrow 10+\left(n-1\right)\times \left(-3\right)=-62\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -3\left(n-1\right)=-62-10=-72 \\ \Rightarrow n-1=\frac{72}{3}=24 \end{gathered}$$
$\Rightarrow n=24+1=25$

Thus, the given AP contains 25 terms.

∴ Middle term of the given AP

= $\left(\frac{25+1}{2}\right)$th term

= 13th term

= 10 + (13 − 1) × (−3)

= 10 − 36

= −26

Hence, the middle term of the given AP is −26.