Question

First three terms of an AP are 2x – 1, x + 1 and x + 3. The value of x is

Answer 1

Question:

First three terms of an AP are 2x – 1, x + 1 and x + 3. Find The value of x.

Solution :

Given AP : 2x – 1, x + 1 ,x + 3

Common difference

$\sf{x+1-(2x-1)=x+3-(x+1)}$

$\sf{x+1-2x+1=x+3-x-1}$

$\sf{2-x=2}$

$\sf{x=2-2}$

$\sf{x=0}$

Therefore, The value of x=0

$\rule{200}2$

More About Arithmetic Progression:

•Genral term of an Ap

$\sf a_{n} = a + (n - 1)d$

•Sum of first n terms of an AP

$\sf \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$

Answer 2

Given :-

First three terms of the AP sequence are :-

2x-1 , X+1 and X+3 .

To Find :-

Value of x .

Solution :-

We are given first 3 terms of a A.P sequence and we know that the common difference between every two terms is constant .

So using this property :-

$\sf{ \implies \: a _2 - a_1 = a_3 - a _2}$

$\sf{ \implies \: X+1 -(2x-1) = (X+3)-(X+1)}$

$\sf{ \implies \: x + 1 - 2x + 1 = x + 3 - x - 1}$

$\sf{ \implies \: -x + 2 = \cancel{x} + 3 \cancel{-x} -1 }$

$\sf{ \implies \: 2 - 2 = x \rightarrow \: 0}$

Hence Value of X is 0

A.P = Arithmetic progression.

General formula :-

$\sf{\implies T_n = a_1 + (n-1) d }$

Sum of n terms :-

$\sf{\implies S_n = \frac{n}{2} [ 2a + (n-1) d] }$

When last term is given :-

$\sf{\implies S_n = \frac{n}{2} [ a + l]}$