Question

if (x-1),(x+1) and (2x+3) are in a. p then the value of x is​

Answer 1

Answer:

x=0

Step-by-step explanation:

(x-1),(x+1),(2x+3)

let a1,a2,a3 are the terms

a2-a1=a3-a2

x+1-(x-1)=2x+3-(x+1)

x+1-x+1=2x+3-x-1

2=x+2

x=0

Answer 2

Step-by-step explanation:

Given:-

(x-1),(x+1) and (2x+3) are in AP

To find:-

Find the value of x?

Solution:-

Method-1:-

Given that

(x-1),(x+1) and (2x+3) are in the AP

Since they are in the AP the common difference is same throughout the series is same.

=> (x+1)-(x-1) = (2x+3)-(x+1)

=> x+1-x+1 = 2x+3-x-1

=> (x-x)+(1+1) = (2x-x)+(3-1)

=> 0+2 = x+2

=> 2 = x+2

=> x+2 = 2

=> x = 2-2

=>x = 0

Method-2:-

We know that

If a , b, c are consecutive terms in the AP then 2b = a+c

We have, a = x-1 , b= x+1 , c= 2x+3

=> 2(x+1) = (x-1) +(2x+3)

=> 2x+2 = x-1+2x+3

=> 2x+2 = 3x+2

=> 2-2 = 3x-2x

=> 0 = x

Therefore, x = 0

Answer:-

The value of x for the given problem is 0

Check:-

If x = 0 then (x-1),(x+1) and (2x+3) becomes

-1 , 1 , 3

Common difference

= 1-(-1)=1+1=2

=3-1 = 2

Since the common difference is same throughout the series. They are in the AP.

Used Concept:-

• The common difference is same throughout the series is same,then they are in the AP.

Used formula:-

• If a , b, c are the consecutive terms in the A. P then 2b = a+c (or) b = (a+c)/2