Which term of the A.P. $-7, -12, -17, -22, …..$ will be $-82$? Is $-100$ any term of the A.P.?

Given:

Given A.P. is $-7, -12, -17, -22, …..$

To do:

We have to find which term of the given A.P. is $-82$ and whether $-100$ is a term of the given A.P.

Solution:

Here,

$a_1=-7, a_2=-12, a_3=-17$

Common difference $d=a_2-a_1=-12-(-7)=-12+7=-5$

We know that,

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

Therefore,

$a_{n}=-7+(n-1)(-5)$

$-82=-7+n(-5)-1(-5)$

$-82+7=-5n+5$

$75+5=5n$

$5n=80$

$n=\frac{80}{5}$

$n=16$

$a_{19}=-7+(19-1)(-5)$

$=-7+18(-5)$

$=-7-90$

$=-97$

$a_{20}=-7+(20-1)(-5)$

$=-7+19(-5)$

$=-7-95$

$=-102$

Hence, $-82$ is the 16th term of the given A.P. and $-100$ is not a term of it.