#### Question

Write the sequence with *n*th term: a_{n} = 9 − 5n

#### Solution

In the given problem, we are given the sequence with the *n*^{th} term `(a_n)`

We need to show that these sequences form an A.P

a_{n} = 9 − 5n

Now, to show that it is an A.P, we will find its few terms by substituting n = 1,2, 3

So

Substituting* n *= 1*, *we get`

`a_1 = 9 - 5(1)`

`a_1 = 4`

Substituting* n *= 2*, *we get

`a_2 = 9 - 5(2)

`a_2 = -1`

Substituting* n *= 3*, *we get

`a_3 = 9 - 5(3)`

`a_3 = -6`

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

Common difference (*d*) = `a_2 - a_1 = a_3 - a_2`

Here

`a_2 - a_1 = -1 - 4

= -5

Also

`a_3 - a_2 = -6 - (-1)`

= -5

Since `a_2 - a_1 = a_3 - a_2`

Hence, the given sequence is an A.P.

Is there an error in this question or solution?

#### APPEARS IN

Solution Write the Sequence with Nth Term: an = 9 − 5n Concept: Arithmetic Progression.