For the A.P.: $-3, -7, -11,…,$ can we find $a_{30} – a_{20}$ without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.
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 $-40$.
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