four numbers in AP such that sum is 20 and sum of whose square is 130
Answers
Correct Question
→ If four numbers in AP such that sum is 20 and sum of whose square is 120
Solution
If four terms are in A.P, then they will be in (a + 3d, a + d, a - d and a - 3d) order.
→ a + 3d + a + d + a - d + a - 3d = 20
→ 4a = 20
→ a = 5
→ (a + 3d)² + (a + d)² + (a - d)² + (a - 3d)² = 120
→ a² + 9d² + 6ad + a² + d² + 2ad + a² + d² - 2ad + a² + 9d² - 6ad = 120
→ 4a² + 20d² = 120
→ 4(5)² + 20d² = 120
→ 20d² = 120 - 100
→ d² = 20/20
→ d = √1 = ± 1
Hence we get an A.P as
→ 8, 6, 4, 2 or 2, 4, 6, 8
Four numbers in AP such that sum is 20 and sum of whose square is 120.
Solution:
» Four numbers in AP such that sum is 20.
• Let four numbers be (a - 3d), (a - d), (a + d) and (a + 3d)
A.T.Q.
=> a - 3d + a - d + a + d + a + 3d = 20 ______ (eq A)
=> 4a = 20
=> a = 20/4
=> a = 5 _______ (eq 1)
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» Sum of whose (let numbers) square is 120.
A.T.Q.
=> (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120
We know that
(a + b)² = a² + b² + 2ab
=> a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 120
=> 4a² + 20d² = 120
=> 4(5)² + 20d² = 120
=> 4 × 25 + 20d² = 120
=> 100 + 20d² = 120
=> 20d² = 20
=> d² = 20/20
=> d² = 1
=> d = ± 1 ______ (eq 2)
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A.P.
» If a = 5 and d = + 1
• a - 3d = 5 - 3(1) = 5 - 3 = 2
• a - d = 5 - 1 = 4
• a + d = 5 + 1 = 6
• a + 3d = 5 + 3(1) = 5 + 3 = 8
» If a = 5 and d = - 1
• a - 3d = 5 - 3(-1) = 5 + 3 = 8
• a - d = 5 - (-1) = 5 + 1 = 6
• a + d = 5 + (-1) = 4
• a + 3d = 5 + 3(-1) = 5 - 3 = 2
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✡ From (eq A)
a - 3d + a - d + a + d + a + 3d = 20
From above calculations we have a = 5 and d = ± 1
Take a = 5 and d = +1 and put it in (eq A)
5 - 3(1) + 5 - 1 + 5 + 1 + 5 + 3(1) = 20
5 - 3 + 5 + 5 + 5 + 3 = 20
20 = 20
• Similarly if we take a = 5 and d = - 1 we get the same result.
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