Question

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

Answer 1

hey mate ✋✋

here's your answer ⏩⏩

Let the four numbers be (a-3d), (a-d), (a+d) & (a+3d).

Now A.T.Q

1. Their sum
(a-3d)+(a-d)+(a+d)+(a+3d)=28
4a = 28
a=7.

2.Their product
(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216
4a²+20d²=216
a² + 5d² = 54
Putting the value of a
49 + 5d² = 54
5d² = 5
d² = 1
d=±1.

Hence, the numbers are 4, 6, 8 & 10.

$\huge{creamiepie}$

$<marquee>If it helped u please mark my answer as BRAINLIEST</marquee>$

#bebrainly

Answer 2

Hope u like my process
=====================

Let the four numbers be

=> (a -3d)

=>( a -d )

=> (a + d)

=> (a +3d)

where 2d is the difference
______________________
Now..
-----------
=> (a-3d)+(a-d)+(a+d)+(a+3d)= 28

=> 4a = 28

$=> \bf \: a = \underline7 \: \: .........(1)$

__________________
Again
----------
=>(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

=> 20d² = 216 -4(7²)

=> 20d² = 216 - 196=20

=> d²= 1

$=> \bf d = \underline1...........(2)$

__________________,______

Hence the numbers are

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

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

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

=>(a+3d) = (7+3) =10
__________________________
Hope this is ur required answer .

Proud to help u