Question

The nth term of an arithmetic progression (A.P.) is (3n + 1) (i) The first three terms of this A. P. are​

Answer 1

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....

Answer 2

Arithmetic Progression

Let ${a}$ and ${d}$ be real numbers. Then the numbers of the form ${a}$, ${a\:+\:d}$, ${a\:+\:2d}$,... 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,

• $\sf{t_n\:=\:3n\:+\:1}$

Substituting values for $\bold{'n'}$,

When $\sf{n\:=\:1},$

$\implies\sf{t_n\:=\:3n\:+\:1}$

$\implies\sf{t_1\:=\:3(1)\:+\:1}$

$\implies\sf{t_1\:=\:3\:+\:1}$

$\implies\sf{t_1\:=\:4}$

When $\sf{n\:=\:2},$

$\implies\sf{t_n\:=\:3n\:+\:1}$

$\implies\sf{t_2\:=\:3(2)\:+\:1}$

$\implies\sf{t_2\:=\:6\:+\:1}$

$\implies\sf{t_2\:=\:7}$

When $\sf{n\:=\:3},$

$\implies\sf{t_n\:=\:3n\:+\:1}$

$\implies\sf{t_3\:=\:3(3)\:+\:1}$

$\implies\sf{t_3\:=\:9\:+\:1}$

$\implies\sf{t_3\:=\:10}$

Hence, the three term in A.P are 4, 7, 10...