Write the sequence with nth term:$a_n=6-n$Show that all of the above sequences form A.P.


Given:

$a_n=6-n$

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 $a_n=6-n$.

Therefore,

$a_1=6-1$

$=5$

$a_2=6-2$

$=4$

$a_3=6-3$

$=3$

$a_4=6-4$

$=2$

The sequence formed is $5, 4, 3, 2,.....$.

For the given sequence to form an A.P., the difference between any two consecutive terms should be equal.

Here,

$d=a_2-a_1=4-5=-1$

$d=a_3-a_2=3-4=-1$

$d=a_4-a_3=2-3=-1$

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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