Question

1. The sum of 4 numbers in AP is 32 and
the sum of their squares is 276. Find the
numbers.
​

Answer 1

Answer:

5, 7, 9, 11

Step-by-step explanation:

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

This is an AP which has common difference of 2d.

Sum = 32

a-3d + a-d + a+d + a+3d = 32

4a = 32

∴ a = 8

Sum of squares = 276

(a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 276

a² - 6ad + 9d² + a² - 2ad + d² + a² + 2ad + d² + a² + 6ad + 9d² = 276

4a² + 20d² = 276

a² + 5d² = 69

Substituting a = 8

8² + 5d² = 69

64 + 5d² = 69

5d² = 5

d² = 1

∴ d = ± 1

Numbers are - 5, 7, 9, 11 or 11, 9, 7, 5

Hope this helps! :)

Please mark as brainliest!

Answer 2

Let us take the four numbers in A.P. to be a−3d,a−d,a+d,a+3d.

Thus, their sum = 4a=32

⇒a=8.

The sum of their squares is 2(a

2

+d

2

)+2(a

2

+9d

2

)=4a

2

+20d

2

=276

⇒a

2

+5d

2

=69

⇒5d

2

=69−64=5, implying d to be 1.

The numbers are thus 5,7,9,11.