Question

the sum of first three terms in arithmetic progression is 24 and the sum of their squares is 224 find the first three terms of this arithmetic progression

Answer 1

Hey,

Let the three terms of an A.P is (a+d), a, (a-d).
:. a+d+a+a-d = 24
=> 3a = 24
=>a = 24/3
=> a = 8............. (1 )

And,
(a+d)^2+a^2+(a-d)^2 = 224
=> a^2+d^2+2ad+a^2+a^2+d^2-2ad=224
=> 3a^2+2b^2 = 224............( 2 )

from equation 1 put the value of a in equation 2 ,we get:
3*8^2 +2d^2 = 224
3*64+2d^2 = 224
192 + 2d^2 = 224
2d^2 = 224-192
=>2d^2= 32
d^2 = 32/2
ď^2 = 16
d = 4

:. the first three terms of an A. P are
a+d = 8+4 = 12
a = 8
a-d = 8-4 = 4... ANSWER

HOPE IT HELPS YOU:-))

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