Show that the sequence defined by an = 3n2 − 5 is not an A.P.

Solution:

In the given problem, we need to show that the given sequence is not an A.P

Here,

$a_{n}=3 n^{2}-5$

Now, first we will find its few terms by substituting $n=1,2,3,4,5$

 

So,

Substituting n = 1, we get

$a_{1}=3(1)^{2}-5$

 

$a_{1}=-2$

Substituting $n=2$, we get

$a_{2}=3(2)^{2}-5$

 

$a_{2}=7$

Substituting n = 3, we get

$a_{3}=3(3)^{2}-5$

 

$a_{3}=22$

Substituting n = 4, we get

$a_{4}=3(4)^{2}-5$

 

$a_{4}=43$

Substituting n = 5, we get

$a_{5}=3(5)^{2}-5$

 

$a_{5}=70$

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

We find the common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$

Thus,

$a_{2}-a_{1}=7-(-2)$

$=9$

Also,

$a_{3}-a_{2}=22-7$

$=15$

So, $a_{2}-a_{1} \neq a_{3}-a_{2}$

Hence, the given sequence is not an A.P.