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Write the sequence with nth term:$a_n=3+4n$Show that all of the above sequences form A.P.
Given:
$a_n=3+4n$
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 $a_n=3+4n$.
Therefore,
$a_1=3+4(1)$
$=3+4$
$=7$
$a_2=3+4(2)$
$=3+8$
$=11$
$a_3=3+4(3)$
$=3+12$
$=15$
$a_4=3+4(4)$
$=3+16$
$=19$
The sequence formed is $7, 11, 15, 19,.....$.
For the given sequence to form an A.P., the difference between any two consecutive terms should be equal.
Here,
$d=a_2-a_1=11-7=4$
$d=a_3-a_2=15-11=4$
$d=a_4-a_3=19-15=4$
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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