Question

Show that

$(a - b) {}^{2} , \: (a + b) {}^{2} and \: (a + b) {}^{2}$

are in AP.

Class 10

Arithmetic Progressions

Answer 1

Hey there !!


➡ Show that :-

( a - b )² , ( a² + b² ) and ( a + b )² are in AP.


➡ Solution:-

The given numbers are ( a - b )² , ( a² + b² ) and ( a + b )².

▶ Now,

= ( a² + b² ) - ( a - b )².

= a² + b² - ( a² - 2ab + b² ).

= a² + b² - a² + 2ab - b².

= 2ab.

And,

= ( a + b )² - ( a² + b² ).

= a² + b² + 2ab - a² - b².

= 2ab.


▶ So, ( a² + b² ) - ( a - b )² = ( a + b )² - ( a² + b² ) = 2ab ( constant ).


▶ Since each term is differ by its preceding terms by a constant, therefore the numbers are in AP.


✔✔ Hence, it is proved ✅✅.

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THANKS

#BeBrainly.

Answer 2

HEY THERE!!!

Correct Question:-

Show that (a-b)², (a²+b²)and (a+b)² are in Arithmetic Progression.

Method Of Solution:-

Let to be T1=(a-b)²

Let to be T2=a²+b²

Let to be T3=(a+b)²

To Find Common Difference must be Subtractaction T1 From T2.

According to the Question;-

T2-T1

=> (a²+b²)-(a-b)²

=> a²+b²-(a²+b²-2ab)

=> a²+b²-a²-b²+2ab

=> 2ab

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Again, Follow the Following Question as per as First Solution;-

*To find Common Difference of Arithmetic Sequence or Progression*

T3-T2

=> (a+b)²-(a²+b²)

=> a²+b²+2ab -a²-b²

=> 2ab

Hence, It's are in Arithmetic Sequence or Progression

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Here, Both Common difference are exactly Same,Sowe considered on Formula of Arithmetic Sequence or Progression;-

If common Difference are same then it's considered to be Arithmetic Sequence are in the Form of arithmetic progression.

Thank you!!