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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Updated on: 10-Oct-2022

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