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Write the sequence with nth term:$a_n=9-5n$Show that all of the above sequences form A.P.
Given:
$a_n=9-5n$
To do:
We have to write the sequence and show that it forms an A.P.
Solution:
To find the given sequence we have to substitute $n=1, 2, 3.....$ in $a_n=9-5n$.
Therefore,
$a_1=9-5(1)$
$=9-5$
$=4$
$a_2=9-5(2)$
$=9-10$
$=-1$
$a_3=9-5(3)$
$=9-15$
$=-6$
$a_4=9-5(4)$
$=9-20$
$=-11$
The sequence formed is $4, -1, -6, -11,.....$.
For the given sequence to form an A.P., the difference between any two consecutive terms should be equal.
Here,
$d=a_2-a_1=-1-4=-5$
$d=a_3-a_2=-6-(-1)=-6+1=-5$
$d=a_4-a_3=-11-(-6)=-11+6=-5$
This implies,
$a_2-a_1=a_3-a_2=a_4-a_3=d$
Hence, the given sequence forms an A.P.
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