Question

Write the AP whose ${\text{n}}^{\text{th}}$ term is given by ${a_n} = 9 - 5n$.

Answer 1

We are given the ${\text{n}}^{\text{th}}$ term of an arithmetic progression and we get the terms. Since the ${\text{n}}^{\text{th}}$ term can be any term of the sequence, using this logic we’ll find the AP by substituting n = 1,2,3…and so on.

Complete step by step solution:
We are given the ${\text{n}}^{\text{th}}$ term of an AP to be ${a_n} = 9 - 5n$
Substituting n = 1 we get the first term of the arithmetic progression
$\Rightarrow {a_1} = 9 - 5(1) = 9 - 5 = 4$
Hence the first term of the arithmetic progression is 4.
We get the next term by substituting n = 2
$\Rightarrow {a_2} = 9 - 5(2) = 9 - 10 = - 1$
The other terms of the AP is given by substituting n = 3 , 4 , ……
$\Rightarrow {a_3} = 9 - 5(3) = 9 - 15 = - 6 \\\ \Rightarrow {a_4} = 9 - 5(4) = 9 - 20 = - 11 \\\$

Therefore the resulting AP is $4 , -1 , -6 , -11,…$

Note:
An arithmetic progression (AP), also called an arithmetic sequence, is a sequence of numbers which differ from each other by a common difference.
Here in the above problem the common difference is found by $d = {a_2} - {a_1}$
Hence , $d = - 1 - 4 = - 5$