Question

the sum of the first three is an arithmetic progression is 24 and sum of their squares is 224 find 3 find first 3 terms of automatic progression​

Answer 1

S3=24

let the terms be a-d,a,a+d

A/Q: a-d+a+a+d=24

3a=24

a=8

Again,(a-d)a(a+d)=224

(8-d)8(8+d)=224

64-d^=224/8

64-d^=28

d^=36

d=6

so , the terms are : 8-6,8,8+6 =2,8,14

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Answer 2

Step-by-step explanation:

Let the first term of the A.P be : a

Let the common difference of the A.P be : d

$\longrightarrow$  Second term of the A.P will be : a + d

$\longrightarrow$  Third term of the A.P will be : a + 2d

Given : Sum of first three terms of the A.P is 24

$\longrightarrow$  a + (a + d) + (a + 2d) = 24

$\longrightarrow$  3a + 3d = 24

$\longrightarrow$  3(a + d) = 24

$\longrightarrow$  a + d = 8

$\longrightarrow$  a = 8 - d

Given : Sum of squares of the first three terms of the A.P is 224

$\longrightarrow$  a² + (a + d)² + (a + 2d)² = 224

$\longrightarrow$  a² + a² + d² + 2ad + a² + 4d² + 4ad = 224

$\longrightarrow$  3a² + 5d² + 6ad = 224

Substituting the value of a = (8 - d) in the above equation, We get :

$\longrightarrow$  3(8 - d)² + 5d² + 6d(8 - d) = 224

$\longrightarrow$  3(64 + d² - 16d) + 5d² + 48d - 6d² = 224

$\longrightarrow$  192 + 3d² - 48d + 5d² + 48d - 6d² = 224

$\longrightarrow$  2d² = 224 - 192

$\longrightarrow$  2d² = 32

$\longrightarrow$  d² = 16

$\longrightarrow$  d = ± 4

Consider : d = 4

$\longrightarrow$  a = (8 - d) = (8 - 4) = 4

$\longrightarrow$  second term : (a + d) = (4 + 4) = 8

$\longrightarrow$  Third term : (a + 2d) = (4 + 8) = 12

In this case : The First three terms of the A.P are 4 , 8 , 12

Consider : d = -4

$\longrightarrow$  a = (8 - d) = (8 + 4) = 12

$\longrightarrow$  second term : (a + d) = (12 - 4) = 8

$\longrightarrow$  Third term : (a + 2d) = (12 - 8) = 4

In this case : The First three terms of the A.P are 12 , 8 , 4