Find the middle term of the AP 10, 7, 4, ..., (−62).

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,

$a_{n}=-62$

$\Rightarrow 10+(n-1) \times(-3)=-62 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow-3(n-1)=-62-10=-72$

$\Rightarrow n-1=\frac{72}{3}=24$

$\Rightarrow n=24+1=25$

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.