In an A.P., sum of four consecutive terms is 28 and their sum of their
squares is 276. Find the four numbers.
Answers
Answer:
The four consecutive terms are 1, 5, 9 and 13.
Step-by-step explanation:
Let the four consecutive terms be (a - 3d), (a - d), (a + d) and (a + 3d)
Given that, The sum of these four terms is 28.
So,
a - 3d + a - d + a + d + a + 3d = 28
4a = 28
a = 7
Also, Given that sum of the square of these terms is 276
=> (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 276
=> a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 276
=> 4a² + 20d² = 276
=> 4(a² + 5d²) = 276
=> a² + 5d² = 69
=> (7)² + 5d² = 69
=> 49 + 5d² = 69
=> 5d² = 69 - 49
=> 5d² = 20
=> d² = 4
=> d = 2
The terms are:
(a - 3d) = 7 - 3(2)
= 7 - 6
= 1
(a - d) = 7 - 2
= 5
(a + d) = 7 + 2
= 9
(a + 3d) = 7 + 3(2)
= 7 + 6
= 13
Answer:
The four terms of the AP are 1,5,9,13
Step-by-step explanation:
Given,
In an AP with the sum of four consecutive terms = 28 and the sum of their squares = 276
To find,
The four terms of the AP
Recall the formula
(a+b)² +(a-b)² = 2a²+2b² ----------------(A)
Solution:
Let the four consecutive terms of the AP be a-3d,a-d,a+d,a+3d
The given,
Sum of the terms = a-3d +a-d +a+d +a+3d = 28 ------------(1)
Sum of their squares = (a-3d)² + (a-d)² +(a+d)² +(a+3d)² = 276 -----(2)
From equation (1)
a-3d +a-d +a+d +a+3d = 28
4a = 28
a = 7
From equation(2)
(a-3d)² + (a-d)² +(a+d)² +(a+3d)² = 276
(a-3d)² + (a+3d)² + (a-d)² +(a+d)² = 276
Applying the formula (A) in the above equation
2a² + 18d² + 2a² + 2d² = 276
4a² + 20d² = 276
Substitute the value of 'a' in the above equation
4×7² +20d² = 276
196 + 20d² = 276
20d² = 276 - 196 = 80
d² = 4
d = ±2
When d =2
a - 3d = 1, a -d = 5, a+d = 9, a+3d = 13
When d =-2
a - 3d = 13, a -d = 9, a+d = 5, a+3d = 1
∴ The four terms of the AP are 13,9,5,1
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