Find $a_{30} - a_{20}$ for the A.P.$a, a + d, a + 2d, a + 3d, …$

Given:

Given A.P. is $a, a + d, a + 2d, a + 3d, …$

To do:

We have to find $a_{30} - a_{20}$.

Solution:

$a_1=a, a_2=a+d, a_3=a+2d$ and $d=a_2-a_1=a+d-(a)=a+d-a=d$

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{30}=a+(30-1)d$

$=a+29(d)$

$=a+29d$

$a_{20}=a+(20-1)d$

$=a+19(d)$

$=a+19d$

This implies,

$a_{30}-a_{20}=a+29d-(a+19d)$

$=a+29d-a-19d$

$=10d$

Hence, $a_{30}-a_{20}$ is $10d$. 

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Updated on: 10-Oct-2022

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