The nth term of an arithmetic progression (A.P.) is (3n + 1) (i) The first three terms of this A. P. are
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The nth term of an arithmetic progression (A.P.) is (3n + 1) (i) The first three terms of this A. P. are
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Step-by-step explanation:
let n=1,
3×1+1=4
n=2,
3×2+1=7
n=3,
3×3+1=10
AP--4,7,10....
Arithmetic Progression
Let [tex]{a}[/tex] and [tex]{d}[/tex] be real numbers. Then the numbers of the form [tex]{a}[/tex], [tex]{a\:+\:d}[/tex], [tex]{a\:+\:2d}[/tex],... is said to be Arithmetic Progression denoted by A. P.
Now, Let's move on finding the solution for our question.
It is given that, The n th term of an arithmetic progression (A.P.) is 3n + 1,
Substituting values for [tex]\bold{'n'}[/tex],
When [tex]\sf{n\:=\:1},[/tex]
[tex]\implies\sf{t_n\:=\:3n\:+\:1}[/tex]
[tex]\implies\sf{t_1\:=\:3(1)\:+\:1}[/tex]
[tex]\implies\sf{t_1\:=\:3\:+\:1}[/tex]
[tex]\implies\sf{t_1\:=\:4}[/tex]
When [tex]\sf{n\:=\:2},[/tex]
[tex]\implies\sf{t_n\:=\:3n\:+\:1}[/tex]
[tex]\implies\sf{t_2\:=\:3(2)\:+\:1}[/tex]
[tex]\implies\sf{t_2\:=\:6\:+\:1}[/tex]
[tex]\implies\sf{t_2\:=\:7}[/tex]
When [tex]\sf{n\:=\:3},[/tex]
[tex]\implies\sf{t_n\:=\:3n\:+\:1}[/tex]
[tex]\implies\sf{t_3\:=\:3(3)\:+\:1}[/tex]
[tex]\implies\sf{t_3\:=\:9\:+\:1}[/tex]
[tex]\implies\sf{t_3\:=\:10}[/tex]
Hence, the three term in A.P are 4, 7, 10...