Question

The general term of a sequence is given by a_n = −4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.

Answer 1

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

Here

${\displaystyle a_n=-4n+15}$

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

${\displaystyle a_1=-4\left(1\right)+15}$

${\displaystyle a_1=11}$

Substituting n = 2, we get

${\displaystyle a_2=-4\left(2\right)+15}$

${\displaystyle a_2=7}$

Substituting n = 3, we get

${\displaystyle a_3=-4\left(3\right)+15}$

${\displaystyle a_3=3}$

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

We find the common difference (d) = ${\displaystyle a_2-a_1=a_3-a_2}$

Thus,

${\displaystyle a_2-a_1=7-11}$

= -4

Also

${\displaystyle a_3-a_2=3-7}$

= -4

Since ${\displaystyle a_2-a_1=a_3-a_2}$

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

Now to find its 15th using the formula ${\displaystyle a_n=a+(n-1)d}$

First term (a) = 11

n = 15

Common difference (d) = −4

Substituting the above values in the formula

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

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

${\displaystyle a_{15}=-45}$

Therefore ${\displaystyle a_{15}=-45}$