Question

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

Answer 1

Given series is $10,7,4,-----(-62)$

first term $\left(a\right)=10$

$d=7-10$

$=-3$

therefore,

$-62=10+(n-1)(-3)$

$-3(n-1)=-62-10$

$-3(n-1)=-72$

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

$n-1=24$

$n=24+1=25$

therefore,

mid-terms is $\frac{(25+1)}{2}$

$=\frac{26}{3}=13$

13th terms$=10+(13-1)\times (-3)$

$=10-36$

$=-26$