Question

the nth term of a sequence is 8-5n.show that the sequence is in A.P

Answer 1

Put n=1 in 8-5n
8-5=3
Put n=2
8-10=-2
Put n=3
8-15=-7
Now -7-(-2)=-2-3
-5=-5
Hence d ( common difference is same)
So the sequence given is in A. P.

Answer 2

Answer:

The proof has been shown below.

Step-by-step explanation:

The nth term of the sequence is 8-5n.

Thus, we have

$t_n=8-5n$

Let us find some terms by plugging n=0,1,2,3

For n=0

$t_0=8-5(0)=8$

For n=1

$t_1=8-5(1)=3$

For n=2

$t_2=8-5(2)=-2$

For n=3

$t_3=8-5(3)=-7$

Thus, the first four terms of the sequence is 8, 3, -2, -7

Let us find the difference between two consecutive terms

3 - 8 = -5

-2 - 3 = -5

-7 -(-2) = -5

We can see that the above differences are same. And in case of AP, the difference between two consecutive terms should be equal.

Hence, we can conclude that the given sequence is in A.P