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

Answer:

\displaystyle \text{Let the three terms of the A.P. be }  a-d, a, a+d

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

\displaystyle \Rightarrow 3a = 21

\displaystyle \Rightarrow a = 7

\displaystyle \text{Also }  (a-d)(a+d) - a = 6

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

\displaystyle \Rightarrow 49 - d^2-7 = 6

\displaystyle \Rightarrow d^2 = 36

\displaystyle \Rightarrow d = \pm 6

\displaystyle \text{When }  a = 7, d = 6 , \text{the three terms are }  1, 7, 13

\displaystyle \text{When }  a = 7, d = -6 , \text{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 \displaystyle 27 and the product \displaystyle 548 , find the numbers.

Answer:

\displaystyle \text{Let the three terms of the A.P. be }  a-d, a, a+d

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

\displaystyle \Rightarrow 3a = 27

\displaystyle \Rightarrow a = 9

\displaystyle \text{Also }  (a-d) \cdot a \cdot (a+d) = 648

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

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

\displaystyle \Rightarrow (81-d^2) = 72

\displaystyle \Rightarrow d^2 = 9

\displaystyle \Rightarrow d = \pm 3

\displaystyle \text{When }  a = 9, d = 3 , \text{the three terms are }  6, 9, 12

\displaystyle \text{When }  a = 9, d = -3 , \text{the three terms are }  12, 9, 6

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

Answer:

\displaystyle \text{Let the four numbers be }  a-3d, a-d, a+d, a+3d

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

\displaystyle \Rightarrow 4a = 50

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

\displaystyle \text{Also }  a+3d = 4 ( a - 3d)

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

\displaystyle \Rightarrow 15 d = 3a

\displaystyle \Rightarrow a = 5d

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

\displaystyle \therefore the terms are \displaystyle \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) \text{ or } 5, 10, 15, 20

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

Answer:

\displaystyle \text{Let the three terms of the A.P. be }  a-d, a, a+d

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

\displaystyle \Rightarrow 3a = 12

\displaystyle \Rightarrow a = 4

\displaystyle \text{Also }  (a-d)^3 + a^3 + ( a+d)^3 = 288

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

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

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

\displaystyle \Rightarrow 192 + 24d^2 = 288

\displaystyle \Rightarrow 24d^2 = 96

\displaystyle \Rightarrow d= \pm 2

\displaystyle \text{When }  a=4, d = 2 , \text{the three terms are }  2,4, 6

\displaystyle \text{When }  a = 4, d = -2 , \text{the three terms are }  6, 4, 2

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

Answer:

\displaystyle \text{Let the three terms of the A.P. be }  a-d, a, a+d

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

\displaystyle \Rightarrow 3a = 24

\displaystyle \Rightarrow a = 8

\displaystyle \text{Also, }  a(a-d)(a+d) = 440

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

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

\displaystyle \Rightarrow d^2 = 9

\displaystyle \Rightarrow d = \pm 3

\displaystyle \text{When }  a = 8, d = 3 , \text{the three terms are }  5, 8, 11

\displaystyle \text{When }  a = 8, d = -3 , \text{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 \displaystyle 10^{\circ} . Find the angles.

Answer:

\displaystyle \text{Let the angles be }  A, (A+d), (A+2d), (A+3d)

\displaystyle \text{Given }  d = 10^{\circ}

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

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

\displaystyle \Rightarrow A = 75^{\circ}

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