Question

four numbers in AP such that sum is 20 and sum of whose square is 130

Answer 1

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

Answer 2

$\textbf{Correct\: Question:}$

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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$\textbf{A.P. is 2, 4, 6, 8 and 8, 6, 4, 2}$

___________ $\bold{[ANSWER]}$

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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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