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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Updated on: 10-Oct-2022

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