Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.
Answers
(a+d) & (a+3d).
Now A.T.Q....
1. Their sum
(a-3d)+(a-d)+(a+d)+(a+3d)=28
4a=28
a=7...
2.Their product
(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=216
4a²+20d²=216
a²+5d²=54
Putting the value of a
49+5d²=54
5d²=5
d²=1
d=±1...
Hence , the numbers are 4, 6., 8 & 10.
DEFINE a AND d:
Let the first term be a
The other terms are (a + d), (a + 2d) and (a + 3d)
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FIND a IN TERM OF d:
The sum is 28
a + (a + d) + (a + 2d) + (a + 3d) = 28
Expand the terms:
a + a + d + a + 2d + a + 3d = 28
Combine like terms
4a + 6d = 28
Divide by 2 through:
2a + 3d = 14
Subtract 3d from both sides:
2a = 14 - 3d
Divide by 2 on both sides:
a = 7 - 3/2 d ---------------------- [ 1 ]
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SOLVE d:
The sum of the squares is 216
a² + (a + d)² + (a + 2d)² + (a + 3d)² = 216
Sub [1 ] into it:
( 7 - 3/2 d )² + ( 7 - 3/2 d + d)² + ( 7 - 3/2 d + 2d)² + ( 7 - 3/2 d + 3d)² = 216
Simply the terms within the bracket:
( 7 - 3/2 d )² + ( 7 - 1/2 d )² + ( 7 + 1/2 d)² + ( 7 + 3/2 d )² = 216
Expand the terms:
7² - 21d + 9/4d² + 7² - 7d + 1/4 d² + 7² + 7d + 1/4 d² + 7² + 21d + 9/4 d² = 216
Combine like terms:
196 + 5d² = 216
Subtract 196
5d² = 20
Divide both sides by 5:
d² = 4
Square root both sides:
d = 2
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SOLVE a:
a = 7 - 3/2 d
a = 7 - 3/2 (2)
a = 4
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FIND THE TERMS:
First term = a = 4
Second term = a + d = 4 + 2 = 6
Third term = a + 2d = 4 + 2(2) = 8
Fourth term = a + 3d = 4 + 3(2) = 10
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Answer: The 4 terms are 4, 6, 8 and 10.
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