The sum of first three terms in an arithmetic progression is 24 and the sum of their squares is 224.Find the first three terms of this arithmetic progression.
Answers
:. 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
Step-by-step explanation:
Let the first term of the A.P be : a
Let the common difference of the A.P be : d
Second term of the A.P will be : a + d
Third term of the A.P will be : a + 2d
Given : Sum of first three terms of the A.P is 24
a + (a + d) + (a + 2d) = 24
3a + 3d = 24
3(a + d) = 24
a + d = 8
a = 8 - d
Given : Sum of squares of the first three terms of the A.P is 224
a² + (a + d)² + (a + 2d)² = 224
a² + a² + d² + 2ad + a² + 4d² + 4ad = 224
3a² + 5d² + 6ad = 224
Substituting the value of a = (8 - d) in the above equation, We get :
3(8 - d)² + 5d² + 6d(8 - d) = 224
3(64 + d² - 16d) + 5d² + 48d - 6d² = 224
192 + 3d² - 48d + 5d² + 48d - 6d² = 224
2d² = 224 - 192
2d² = 32
d² = 16
d = ± 4
Consider : d = 4
a = (8 - d) = (8 - 4) = 4
second term : (a + d) = (4 + 4) = 8
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
a = (8 - d) = (8 + 4) = 12
second term : (a + d) = (12 - 4) = 8
Third term : (a + 2d) = (12 - 8) = 4
In this case : The First three terms of the A.P are 12 , 8 , 4