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
Answer:
Let the roots of A.P .be (a-d) ,a,(a+d)
Now the sum of 1st three terms is 24 that means
a-d+a+a+d=24
3a=24
a=8
we got the value of a=8
Now,
squaring the terms
(a-d) 2 +(a) 2+ (a-d) 2 =224
where 2 are square not in multiplication
a2+d2-2ad+a2+a2+d2+2ad=224
3a 2 +2d 2 =224
3×8×8+2d2=224
2d2=224–192
2d2=32
d2=16
d=4
Therefore a-d,a,a+d are
8–4,8,8+4
4,8,12 ←This are the values of 1st three 3terms
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