Question 1: The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6 , find three terms.

Answer:

Let the three terms of the A.P. be a-d, a, a+d

\therefore (a-d) + a + ( a+d) = 21

\Rightarrow 3a = 21

\Rightarrow a = 7

Also (a-d)(a+d) - a = 6

\Rightarrow a^2 - d^2 - a = 6

\Rightarrow 49 - d^2-7 = 6

\Rightarrow d^2 = 36

\Rightarrow d = \pm 6

When a = 7, d = 6  ,  the three terms are 1, 7, 13

When a = 7, d = -6  , the three terms are 13, 7, 1

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Question 2: Three numbers are in A.P. If the sum of these numbers be 27 and the product 548 , find the numbers.

Answer:

Let the three terms of the A.P. be a-d, a, a+d

\therefore (a-d) + a + ( a+d) = 27

\Rightarrow 3a = 27

\Rightarrow a = 9

Also (a-d) \cdot a \cdot (a+d) = 648

\Rightarrow a(a^2 - d^2) = 648

\Rightarrow 9(81-d^2) = 648

\Rightarrow (81-d^2) = 72

\Rightarrow d^2 = 9

\Rightarrow d = \pm 3

When a = 9, d = 3 ,  the three terms are 6, 9, 12

When a = 9, d = -3 , the three terms are 12, 9, 6

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Question 3: Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Answer:

Let the four numbers be a-3d, a-d, a+d, a+3d

\therefore (a-3d) + ( a-d) + (a+d) + ( a+3d) = 50

\Rightarrow 4a = 50

\Rightarrow a = \frac{25}{2}

Also a+3d = 4 ( a - 3d)

\Rightarrow a+3d = 4a - 12 d

\Rightarrow 15 d = 3a

\Rightarrow a = 5d

\Rightarrow d = \frac{25}{2}   \times \frac{1}{5}   = \frac{5}{2}

\therefore the terms are \Big( \frac{25}{2}- 3 \times \frac{5}{2}  \Big), \Big( \frac{25}{2}- \frac{5}{2}  \Big) , \Big( \frac{25}{2}+ \frac{5}{2}  \Big) , \Big( \frac{25}{2}+ 3 \times \frac{5}{2}  \Big) or 5, 10, 15, 20

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Question 4: The sum of three numbers in A.P. is 12 and the sum of their cubes is 288 . Find the numbers.

Answer:

Let the three terms of the A.P. be a-d, a, a+d

\therefore (a-d) + a + ( a+d) = 12

\Rightarrow 3a = 12

\Rightarrow a = 4

Also (a-d)^3 + a^3 + ( a+d)^3 = 288

\Rightarrow a^3 - d^3 -3a^2d + 3ad^2 + a^3 +a^3 +d^3 + 3a^2d+ 3ad^2 = 288

\Rightarrow 3a^3 + 6ad^2 = 288

\Rightarrow 3(4)^3 + 6(4) d^2 = 288

\Rightarrow 192 + 24d^2 = 288

\Rightarrow 24d^2 = 96

\Rightarrow d= \pm 2

When a=4, d = 2 ,  the three terms are 2,4, 6

When a = 4, d = -2 , the three terms are 6, 4, 2

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Question 5: If the sum of three numbers in A.P. is 24 and their product 440 , find the numbers.

Answer:

Let the three terms of the A.P. be a-d, a, a+d

\therefore (a-d) + a + ( a+d) = 24

\Rightarrow 3a = 24

\Rightarrow a = 8

Also, a(a-d)(a+d) = 440

\Rightarrow a(a^2 - d^2) = 440

\Rightarrow 8(64-d^2) = 440

\Rightarrow d^2 = 9

\Rightarrow d = \pm 3

When a = 8, d = 3 ,  the three terms are 5, 8, 11

When a = 8, d = -3 , the three terms are 11, 8, 5

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Question 6: The angles of a quadrilateral are in A.P. whose common difference is 10^{\circ} . Find the angles.

Answer:

Let the angles be A, (A+d), (A+2d), (A+3d)

Given d = 10^{\circ}

Sum of the angles A+ (A+d)+ (A+2d)+ (A+3d) = 360^{\circ}

\Rightarrow 4A + 60^{\circ} = 360^{\circ}

\Rightarrow A = 75^{\circ}

Therefore the angles are 75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ} .