Question

find four numbers in a.p. whose sum is 20 and the sum of whose squares is 120​

Answer 1

Answer:

2, 4, 6, 8

Step-by-step explanation:

Let the the numbers be a-3d , a-d, a+d, a+3d

You can see that the common difference is 2d

given their sum is 20, therefore,

a-3d+a-d+a+d+a+3d = 20

the terms with d get canceled out

4a = 20

a = 5

now the sum of their squares is 120

(a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120

if you expand the squares, you will get

$4a^{2} +20d^{2} =120\\ a^2 + 5d^2 = 30$

put a=4

$5^2+5d^2=30 \\ 5d^2=5 \\ d^2=1 \\ d=1$

therefore, numbers are 2,4,6,8

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Answer 2

Answer:

Given:

• $Numbers \ are \ in \ AP$
• $Sum = 20$
• $Sum \ of \ squares = 120$

Solution:

Let the numbers are (a - 3d) , (a - d) , (a + d) , (a + 3d)

According to question:

$(a-3d) + (a-d) + (a + d)+(a+3d) = 20$

$\implies a-3d+a-d+a+d+a+3d = 20$

$\implies 4a = 20$

$\implies a = 5$

Now :

$(a-3d)^2 + (a-d)^2 + (a + d)^2 + (a+3d)^2= 120$

$(5-3d)^2 + (5-d)^2 + (5+d)^2 + (5+3d)^2 = 120$

$25 + 9d^2 - 30d + 25 + d^2 - 10d + 25 + d^2 + 10d + 25 + 9d^2 + 30d = 120$

$100 + 20d = 120$

$\implies d = 1$

Hence the numbers are 2 , 4 , 6 , 8