Question

Q. The sum of four members in A.P is 26 and sum of their squares is 214.Find the numbers.

Answer 1

Hi Mate !! 

Let the terms of AP be 

( a - 3d ) ( a - d ) ( a + d ) ( a + 3d ) 

Where , a is first term
and d is common difference

• Sum is 26

a - 3d + a - d + a + d + a + 3d = 26

4a = 26

a = 26/4

a = 13/2

a = 6.5

• Sum of their square is 214

[ using identity :- ( a - b )² = a² + b² - 2ab 
( x + y ) = x² + y² + 2xy ]
( a - 3d )² + ( a - d )² + ( a + d )² + ( a + 3d )² = 214

a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 214

4a² + 20d² = 214 

Putting value of a 

4 × ( 6.5 )² + 20d² =214 

169 + 20d² = 214

20d² = 214 - 169

20d² = 45

d² = 45/20

d² = 2.25

d = √2.25

d = ± 1.5

So, the nos. are :-

• If a = 6.5 and d = 1.5

( a - 3d ) ( a - d ) ( a + d ) ( a + 3d ) 

2 , 5 , 8 , 11

• If a = 6.5 and d = 1.5

( a - 3d ) ( a - d ) ( a + d ) ( a + 3d ) 


11 , 8 , 5 , 2

Answer 2

Hey !!

Here's your answer.. ⬇⬇

Suppose the four numbers are ( a-3d ), ( a-d ), ( a+d ), ( a+3d ).

:- Sum of their is 26.

a - 3d + a - d + a + d + a + 3d = 26
a + a + a + a = 26
4a = 26
a = 26/4
a = 6.5

:- Sum of their square is 214.

( a-3d )^2 + ( a-d )^2 + ( a+d )^2 + ( a+3d )^2 = 214

a^2 - 6ad + 9d^2 + a^2 - 2ad + d^2 + a^2 + 2ad + d^2 + a^2 + 6ad + 9d^2 = 214

4a^2 + 20d^2 = 214
4 ( 6.5 )^2 + 20d^2 = 214 ---- ( a = 6.5 )
169 + 20d^2 = 214
20d^2 = 214 - 169
20d^2 = 45
d^2 = 45/20
d^2 = 9/4
d = 3/2
d = 1.5

Now,
( a-3d ) = ( 6.5 - 3( 1.5 )) = 2
( a-d ) = ( 6.5 - 1.5 ) = 5
( a+d ) = ( 6.5 + 1.5 ) = 8
( a+3d ) = ( 6.5 + 3(1.5)) = 11


The four numbers are 2, 5, 8, 11.

HOPE IT HELPS..

THANKS ^-^