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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Updated on: 10-Oct-2022

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