Question

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

Answer 1

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

A.T.Q

sum

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

4a = 28

a=7

product

(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216

4a²+20d²=216

a² + 5d² = 54

Put the value of a

49 + 5d² = 54

5d² = 5

d² = 1

d=±1

the numbers are 4, 6, 8 & 10 when a=7 d=1

Numbers are 10,8,6 & 4 when a=7 and d=-1

@skb

Answer 2

Hey there !!


→ Let the required number be ( a - 3d ), ( a - d ) , ( a + d ) and ( a + 3d ).


▶Now,

A/Q,

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

=> a - 3d + a - d + a + d + a + 3d = 28.

=> 4a = 28.

=> a = 28/4.

=> a = 7.

▶Again,

=> ( a - 3d )² + ( a - d )² + ( a + d )² + ( a + 3d )² = 216.

=> a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 216.

=> 4a² + 20d² = 216.

=> 4 × 7² + 20d² = 216.

=> 4 × 49 + 20d² = 216.

=> 20d² = 216 - 196.

=> d² = 20/20.

=> d = √1.

=> d = ±1.

➡ When taking , a = 7 and d = 1, we get

=> a - 3d = 7 - 3 = 4.

=> a - d = 7 - 1 = 6.

=> a + d = 7 + 1 = 8.

=> a + 3d = 7 + 3 = 10.


➡ When taking , a = 7 and d = -1, we get

=> a - 3d = 7 + 3 = 10.

=> a - d = 7 + 1 = 8.

=> a + d = 7 - 1 = 6.

=> a + 3d = 7 - 3 = 4.


✔✔ Hence, the required number are ( 4, 6, 8, 10 ) or ( 10, 8, 6, and 4 ). ✅✅

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THANKS

#BeBrainly.