Written Instructions:
For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..
For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.
Leave your answers in the simplest form or correct to two decimal places.
| 1) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 2) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 3) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 4) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 5) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 6) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 7) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 8) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 9) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 10) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
Answer:_______________ |
| 1) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 2) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 3) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 4) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 5) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 6) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 7) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 8) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 9) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |
| 10) The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers. Answer: -6,2,10 SOLUTION 1 : Given their sum is 6 ⇒ (a-d) + (a) + (a+d) = 6 ⇒ a - ⇒ 3a = 6 ⇒ a = ⇒ a = 2 Also given that their product is -120 ⇒ (a-d) + (a) + (a+d) = -120 ⇒ a(a2-d2) = -120 ⇒ 2 (22 - d2) = -120 ⇒ (4 - d2) = -60 ⇒ -d2 = -60-4 ⇒ -d2 = -64 ⇒ d2 = 64 = 8 (i) when a = 2, d = 8 The three number are 2 - 8, 2, 2+8 -6, 2, 10. (ii) when a = 2, d = -8 The three number are 2 -(- 8), 2, 2- 8 Three numbers in A.P are -6, 2, 10 |