Math, asked by SƬᏗᏒᏇᏗƦƦᎥᎧƦ, 5 hours ago

The sum of four numbers in arithmetic progression is 16.The square of the last number is the square of the first number plus 48. What is the third number?
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Answers

Answered by pandeyvaishnavi88
69

Answer:

A+A+d+A+2d+a+3d=16

4A+6d=16

2A+3d=8

(A+3d)^2=A^2+48

second equation...

A^2+6dA+9d^2=a^2+48

3d(2A+3d)=48

3d*8=48

d=2

then 2A+6=8

A=1

1,3,5,7

Step-by-step explanation:

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Answered by s1274himendu3564
40

The sum of four numbers in arithmetic progression is 16.The square of the last number is the square of the first number plus 48. What is the third number

The four numbers in the AP can be taken as

a - 3d, a - d, a + d and a + 3d.

Given their sum is 16.

=> 4a = 16 => a = 4.

Also given (4 + 3d)² = (4 - 3d)² + 48

=> 16 + 24d + 9d² = 16 - 24 d + 9d² + 48

=> 48 d = 48 => d = 1.

So the numbers are 1, 3, 5, 7

The third term in the sequence is 5

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