Find $a_{30} - a_{20}$ for the A.P.$-9, -14, -19, -24, …$
Given:
Given A.P. is $-9, -14, -19, -24, …$
To do:
We have to find $a_{30} - a_{20}$.
Solution:
$a_1=-9, a_2=-14, a_3=-19$ and $d=a_2-a_1=-14-(-9)=-14+9=-5$
We know that,
nth term of an A.P. $a_n=a+(n-1)d$
Therefore,
$a_{30}=a+(30-1)d$
$=-9+29(-5)$
$=-9-145$
$=-154$
$a_{20}=a+(20-1)d$
$=-9+19(-5)$
$=-9-95$
$=-104$
This implies,
$a_{30}-a_{20}=-154-(-104)$
$=-154+104$
$=-50$
Hence, $a_{30}-a_{20}$ is $-50$.
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