Question

Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.

Answer 1

Question;-

Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.


Method Of Solution;-


Let to be Required Number of (a-3d),(a-d),(a+d) and (a-3d) . It is in the form of Arithmetic Sequence or Progression Equation!








According to the Question as per as Given in first term of Question-


=> (a-3d)+(a-d)+(a+d)+(a+3d)= 28

=> 4a = 28

=> a=28/4

=> a=7 ( Equation 1)

Considering on Second terms which are Given in Question based ;-


Squaring on terms of Number;-

=>(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216

=>(a-d)²+(a+d)²+(a-3d)²+(a+3d)²=216

=> (2a² +2d²)+ (2a²+ 18d²)=216

=> 4a² +20d² =216

=> 4(7)²+20d=216

=> 20d² = 216 -4(49)

=> 20d² = 216 - 196=20

=> d²= 1 

=> d=±1

Taken , Positive 1



Hence Required terms of Number;


Substitute the value of a and d in Equation;-



=> (a-3d) = (7-3)=4

=>(a-d)=(7-1)= 6

=>(a+d)= (7+1)= 8

=>(a+3d) = (7+3) =10


Hence, Required Number are 6, 8, 10 and 10,8,6.


Thanks!!!

Answer 2

$\huge{hey}$

$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{see \: in \: pic}$

$\sf{hence \: required \: no \: is \: 4..6..8..10.......and so on }$

$\huge \boxed{ \ulcorner{hope \: it \: helps}}$

$\huge{ \mathbb{THANKS}}$...

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