Show that the sequence defined by an = 5n −7 is an A.P,

Solution:

In the given problem, we need to show that the given sequence is an A.P and then find its common difference.

Here,

$a_{n}=5 n-7$

Now, to show that it is an A.P, we will find its few terms by substituting 

So,

Substituting n = 1, we get

$a_{1}=5(1)-7$

$a_{1}=-2$

Substituting n = 2, we get

$a_{2}=5(2)-7$

 

$a_{2}=3$

Substituting n = 3, we get

$a_{3}=5(3)-7$

 

$a_{3}=8$

Substituting n = 4, we get

$a_{4}=5(4)-7$

 

$a_{4}=13$

Substituting n = 5, we get

$a_{5}=5(5)-7$

 

$a_{5}=18$

Further, for the given sequence to be an A.P,

We find the common difference (d)

$d^{\prime}=a_{2}-a_{1}=a_{3}-a_{2}$

Thus,

$a_{2}-a_{1}=3-(-2)$

$=5$

Also,

$a_{3}-a_{2}=8-3$

$=5$

Since $a_{2}-a_{1}=a_{3}-a_{2}$

Hence, the given sequence is an A.P and its common difference is $d=5$.