# For the AP: $-3,-7,-11, \ldots$, can we find directly $a_{30}-a_{20}$ without actually finding $a_{30}$ and $a_{20}$ ? Give reasons for your answer.

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Given:

Given A.P. is $-3, -7, -11,…,$

To do:

We have to find $a_{30} – a_{20}$ without actually finding $a_{30}$ and $a_{20}$.

Solution:

In the given A.P.

$a_1=-3, a_2=-7$

$d=a_2-a_1=-7-(-3)=-7+3=-4$

We know that,

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

Therefore,

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

$=a+29d$

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

$=a+19d$

This implies,

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

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

$=10d$

$=10(-4)$

$=-40$

Yes, we can find the value of $a_{30} – a_{20}$ without actually finding $a_{30}$ and $a_{20}$.

$a_{30}$ is actually  $a_{20}+10d$.

Hence, the value of $a_{30} – a_{20}$ is $-40$.

Updated on 10-Oct-2022 13:27:27