Question

How many terms aare there in the AP 41, 38, 35,..., 8?

Answer 1

$A.P.=41,38,35,......8$
$a=41$

$a_n=8$

$d=a_2-a_1=38-41=-3$

nth term, $a_n=a+(n-1)d$

$8=41+(n-1)(-3)$

$-41+8=(n-1)(-3)$

$-33=(n-1)(-3)$

$\frac{-33}{-3}=n-1$

$11=n-1$

$11+1=n$

$12=n$

therefore, the number of terms in the given A.P. is 12