Question

4 numbers are in AP if their sum is 20 and the sum of their squares is 120 then find the numbers ​

Answer 1

Answer:

Given:

Four numbers are in AP. Their sum is 20 and the sum of their squares is 120

To Find:

We need to find the four numbers.

Solution:

Let the numbers in AP be

a-3d, a-d,a+d,a+3d________(1)

We have been given that the sum of the terms is 20 therefore,

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

4a = 20

a = 20/4

a = 5_________(2)

It is also given that the sum of the squares of the terms is 120 therefore,

(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______(3)

Substituting the value of a in equation 3 from equation 2. We get,

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

=> 100 + 20d^2 = 120

=> 20d^2 = 20

=> d = +1 (or) - 1

Since, AP cannot be negative, so

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

a - 3d = 5 - 3(1) = 5 - 3 = 2

a-d = 5 - 1 = 4

a+d = 1 + 5 = 6

a+3d = 5 + 3(1) = 5 + 3 = 8

Hence, the required four numbers in AP are 2,4,6,8....

Answer 2

Answer:

$\implies$ a - 3d = 5 - 3(1) = 5 - 3 = 2.

$\implies$ a - d = 5 - 1 = 4.

$\implies$ a + d = 5 + 1 = 6.

$\implies$ a + 3d = 5 + 3(1) = 5 + 3 = 8.

$\large\underline\mathrm{Given:-}$

• Four numbers are in AP if their sum is 20 and the sum of their squares is 120.

$\large\underline\mathrm{To \: find}$

• We need the for numbers.

$\large\underline\mathrm{Solution}$

• Let the number in Ap.

a - 3d, a - d, a + d, a + 3d. ....(1)

The sum of the terms is 20.

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

$\implies$ 4a = 20

$\implies$ a = 20/4

$\implies$ a = 5. ....(2)

The sum of the squares of the terms is 120.

$\implies$ (a - 3d)² + (a - d)²+ (a + d)² + (a + 3d)² = 20

$\implies$ (a² + 9d² - 2ad) + (a² + d² - 2ad) + (a² + d² + 2ad) + (a² + 9d² + 6ad) = 120

$\implies$ 4a² - 20d² = 120...(3)

Now, the value of a in equation (3) from equation (2). we get.

$\implies$ 4a² + 20d² = 120

$\implies$ 4(5)² + 20d² = 120

$\implies$ 100 + 20d² = 120

$\implies$ 20d² =120 - 100

$\implies$ 20d² = 20

$\implies$ d² = 1

$\implies$ d = (+1, -1)

Thus, Ap can't be negative;

so,

$\implies$ a = 5

$\implies$ d = 1

$\implies$ a - 3d = 5 - 3(1) = 5 - 3 = 2.

$\implies$ a - d = 5 - 1 = 4.

$\implies$ a + d = 5 + 1 = 6.

$\implies$ a + 3d = 5 + 3(1) = 5 + 3 = 8.