The general term of a sequence is give by an = −4n + 15.

Solution:

In the given problem, we need to find that the given sequence is an A.P or not and then find its $15^{\text {th }}$ term and the common difference.

Here,

$a_{n}=-4 n+15$

Now, to find that it is an A.P or not, we will find its few terms by substituting $n=1,2,3$

So,

Substituting n = 1, we get

$a_{1}=-4(1)+15$

 

$a_{1}=11$

Substituting n = 2, we get

$a_{2}=-4(2)+15$

 

$a_{2}=7$

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,

$a_{2}-a_{1}=7-11$

$=-4$

Also,

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

$=-4$

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

Hence, the given sequence is an A.P and its common difference is $d=-4$

 

Now, to find its $15^{\text {th }}$ using the formula $a_{n}=a+(n-1) d$

First term (a) = 11

n = 15

Common difference (d) = −4

Substituting the above values in the formula

$a_{15}=11+(15-1)(-4)$

$a_{15}=11+(-56)$

$a_{15}=-45$

Therefore, $a_{15}=-45$