Question

find the 4 numbers in an AP whose Sum is 10 and the sum of whose square is 30​

Answer 1

Answer:

Let the four numbers in A.P be a-3d, a-d,a+d,a+3d. ---- (1)

Given that Sum of the terms = 20.

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

4a = 20

a = 5. ---- (2)

Given that sum of squares of the term = 120.

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

= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120

= 4a^2 + 20d^2 = 120

Substitute a = 5 from (2) .

4(5)^2 + 20d^2 = 120

100 + 20d^2 = 120

20d^2 = 20

d = +1 (or) - 1.

Since AP cannot be negative.

Substitute a = 5 and d = 1 in (1), we get

a - 3d, a-d, a+d, a+3d = 2,4,6,8.

Hope this helps!

Answer 2

Answer:

YOUR ANSWER

Step-by-step explanation:

Let the numbers be: a−3d,a−d,a+d,a+3d

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=32

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d)

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d)

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d)

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d)

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =276

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =276

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2 =276−4(8)

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2 =276−4(8) 2

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2 =276−4(8) 2 =20

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2 =276−4(8) 2 =20∴d=1

Let the numbers be: a−3d,a−d,a+d,a+3da−3d+a−d+a+d+a+3d=324a=32∴a=8(a−3d) 2 +(a−d) 2 +(a+d) 2 +(a+3d) 2 =2764a 2 +20d 2 =27620d 2 =276−4(8) 2 =20∴d=1Therefore the 4 numbers are: a−3d,a−d,a+d,a+3d=8−3,8−1,8+1,8+3={5,7,9,11}

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