Page 1 1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP. 3 2 , 4 6 k k 2 k Sol: It is given that and are three consecutive terms of an AP. 4 3 4 k 6 4 6 3 2 4 6 4 3 8 3 8 4 4 12 3 k k k k k k k k k k k 6 2 2 2 k k Hence, the value of k is 3. 2. 1 5 2 , 4 x x 2 x Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP. Sol: It is given that and are in AP. 4 5 2 4 x 1 4 1 5 2 2 4 1 3 3 3 3 3 2 x 6 3 x x x x x x x x 3 x x x x 1 2 Hence, the value of x is 3. Page 2 1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP. 3 2 , 4 6 k k 2 k Sol: It is given that and are three consecutive terms of an AP. 4 3 4 k 6 4 6 3 2 4 6 4 3 8 3 8 4 4 12 3 k k k k k k k k k k k 6 2 2 2 k k Hence, the value of k is 3. 2. 1 5 2 , 4 x x 2 x Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP. Sol: It is given that and are in AP. 4 5 2 4 x 1 4 1 5 2 2 4 1 3 3 3 3 3 2 x 6 3 x x x x x x x x 3 x x x x 1 2 Hence, the value of x is 3. 3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y. Sol: It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP. 3 5 3 1 5 1 3 5 3 5 3 1 5 1 3 5 6 2 4 2 6 4 10 5 y y y y y y y y y y y Hence, the value of y is 5. 4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP. Sol: Since 2 ,2 x x and 2 3 x are in AP, we have 2 2 2 3 2 2 3 5 5 x x x x x x x 5. Show that 2 2 2 , a b a b and 2 2 a b are in AP. Sol: The given numbers are 2 2 2 , a b a b and 2 . a b Now, 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a a b b a b a a b b a b a b a b a a b b a b a b So, 2 2 2 2 2 2 2 a b a b a b a b a b (Constant) Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP. 6. Find the three numbers in AP whose sum is 15 and product is 80. Sol: Let the required numbers be , a d a and . a d Then 15 a d a a d Page 3 1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP. 3 2 , 4 6 k k 2 k Sol: It is given that and are three consecutive terms of an AP. 4 3 4 k 6 4 6 3 2 4 6 4 3 8 3 8 4 4 12 3 k k k k k k k k k k k 6 2 2 2 k k Hence, the value of k is 3. 2. 1 5 2 , 4 x x 2 x Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP. Sol: It is given that and are in AP. 4 5 2 4 x 1 4 1 5 2 2 4 1 3 3 3 3 3 2 x 6 3 x x x x x x x x 3 x x x x 1 2 Hence, the value of x is 3. 3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y. Sol: It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP. 3 5 3 1 5 1 3 5 3 5 3 1 5 1 3 5 6 2 4 2 6 4 10 5 y y y y y y y y y y y Hence, the value of y is 5. 4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP. Sol: Since 2 ,2 x x and 2 3 x are in AP, we have 2 2 2 3 2 2 3 5 5 x x x x x x x 5. Show that 2 2 2 , a b a b and 2 2 a b are in AP. Sol: The given numbers are 2 2 2 , a b a b and 2 . a b Now, 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a a b b a b a a b b a b a b a b a a b b a b a b So, 2 2 2 2 2 2 2 a b a b a b a b a b (Constant) Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP. 6. Find the three numbers in AP whose sum is 15 and product is 80. Sol: Let the required numbers be , a d a and . a d Then 15 a d a a d 3 15 5 a a Also, . . 80 a d a a d 2 2 2 2 80 5 25 80 25 16 9 3 a a d d d d Thus, 5 a and 3 d Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d 7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers. Sol: Let the required numbers be , a d a and . a d Then 3 a d a a d 3 3 1 a a Also, . . 35 a d a a d 2 2 2 2 35 1. 1 35 36 6 a a d d d d Thus, 1 a and 6 d Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d 8. Divide 24 in three parts such that they are in AP and their product is 440. Sol: Let the required parts of 24 be , a d a and a d such that they are in AP. Then 24 a d a a d 3 24 8 a a Also, . . 440 a d a a d 2 2 2 440 8 64 440 a a d d Page 4 1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP. 3 2 , 4 6 k k 2 k Sol: It is given that and are three consecutive terms of an AP. 4 3 4 k 6 4 6 3 2 4 6 4 3 8 3 8 4 4 12 3 k k k k k k k k k k k 6 2 2 2 k k Hence, the value of k is 3. 2. 1 5 2 , 4 x x 2 x Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP. Sol: It is given that and are in AP. 4 5 2 4 x 1 4 1 5 2 2 4 1 3 3 3 3 3 2 x 6 3 x x x x x x x x 3 x x x x 1 2 Hence, the value of x is 3. 3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y. Sol: It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP. 3 5 3 1 5 1 3 5 3 5 3 1 5 1 3 5 6 2 4 2 6 4 10 5 y y y y y y y y y y y Hence, the value of y is 5. 4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP. Sol: Since 2 ,2 x x and 2 3 x are in AP, we have 2 2 2 3 2 2 3 5 5 x x x x x x x 5. Show that 2 2 2 , a b a b and 2 2 a b are in AP. Sol: The given numbers are 2 2 2 , a b a b and 2 . a b Now, 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a a b b a b a a b b a b a b a b a a b b a b a b So, 2 2 2 2 2 2 2 a b a b a b a b a b (Constant) Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP. 6. Find the three numbers in AP whose sum is 15 and product is 80. Sol: Let the required numbers be , a d a and . a d Then 15 a d a a d 3 15 5 a a Also, . . 80 a d a a d 2 2 2 2 80 5 25 80 25 16 9 3 a a d d d d Thus, 5 a and 3 d Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d 7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers. Sol: Let the required numbers be , a d a and . a d Then 3 a d a a d 3 3 1 a a Also, . . 35 a d a a d 2 2 2 2 35 1. 1 35 36 6 a a d d d d Thus, 1 a and 6 d Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d 8. Divide 24 in three parts such that they are in AP and their product is 440. Sol: Let the required parts of 24 be , a d a and a d such that they are in AP. Then 24 a d a a d 3 24 8 a a Also, . . 440 a d a a d 2 2 2 440 8 64 440 a a d d 2 64 55 9 3 d d Thus, 8 a and 3 d Hence, the required parts of 24 are 5,8,11 11,8,5 . o r 9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms Sol: Let the required terms be , a d a and . a d Then 21 a d a a d 3 21 7 a a Also, 2 2 2 165 a d a a d 2 2 2 2 2 3 2 165 3 49 2 165 2 165 147 18 9 3 a d d d d d Thus, 7 a and 3 d Hence, the required terms are 4,7,10 10,7,4 . o r 10. The angles of quadrilateral are in whose AP common difference is 10 . Find the angles. Sol: Let the required angles be 15 , 5 , 5 15 , a a a a n d a as the common difference is 10 (given). Then 15 5 5 15 360 a a a a 4 360 90 a a Hence, the required angles of a quadrilateral are 90 15 , 90 5 , 90 5 90 15 ; 75 ,85 ,95 105 . a n d o r a n d 11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216. Sol: (4, 6, 8, 10) or (10, 8, 6, 4) Page 5 1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP. 3 2 , 4 6 k k 2 k Sol: It is given that and are three consecutive terms of an AP. 4 3 4 k 6 4 6 3 2 4 6 4 3 8 3 8 4 4 12 3 k k k k k k k k k k k 6 2 2 2 k k Hence, the value of k is 3. 2. 1 5 2 , 4 x x 2 x Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP. Sol: It is given that and are in AP. 4 5 2 4 x 1 4 1 5 2 2 4 1 3 3 3 3 3 2 x 6 3 x x x x x x x x 3 x x x x 1 2 Hence, the value of x is 3. 3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y. Sol: It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP. 3 5 3 1 5 1 3 5 3 5 3 1 5 1 3 5 6 2 4 2 6 4 10 5 y y y y y y y y y y y Hence, the value of y is 5. 4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP. Sol: Since 2 ,2 x x and 2 3 x are in AP, we have 2 2 2 3 2 2 3 5 5 x x x x x x x 5. Show that 2 2 2 , a b a b and 2 2 a b are in AP. Sol: The given numbers are 2 2 2 , a b a b and 2 . a b Now, 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a a b b a b a a b b a b a b a b a a b b a b a b So, 2 2 2 2 2 2 2 a b a b a b a b a b (Constant) Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP. 6. Find the three numbers in AP whose sum is 15 and product is 80. Sol: Let the required numbers be , a d a and . a d Then 15 a d a a d 3 15 5 a a Also, . . 80 a d a a d 2 2 2 2 80 5 25 80 25 16 9 3 a a d d d d Thus, 5 a and 3 d Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d 7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers. Sol: Let the required numbers be , a d a and . a d Then 3 a d a a d 3 3 1 a a Also, . . 35 a d a a d 2 2 2 2 35 1. 1 35 36 6 a a d d d d Thus, 1 a and 6 d Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d 8. Divide 24 in three parts such that they are in AP and their product is 440. Sol: Let the required parts of 24 be , a d a and a d such that they are in AP. Then 24 a d a a d 3 24 8 a a Also, . . 440 a d a a d 2 2 2 440 8 64 440 a a d d 2 64 55 9 3 d d Thus, 8 a and 3 d Hence, the required parts of 24 are 5,8,11 11,8,5 . o r 9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms Sol: Let the required terms be , a d a and . a d Then 21 a d a a d 3 21 7 a a Also, 2 2 2 165 a d a a d 2 2 2 2 2 3 2 165 3 49 2 165 2 165 147 18 9 3 a d d d d d Thus, 7 a and 3 d Hence, the required terms are 4,7,10 10,7,4 . o r 10. The angles of quadrilateral are in whose AP common difference is 10 . Find the angles. Sol: Let the required angles be 15 , 5 , 5 15 , a a a a n d a as the common difference is 10 (given). Then 15 5 5 15 360 a a a a 4 360 90 a a Hence, the required angles of a quadrilateral are 90 15 , 90 5 , 90 5 90 15 ; 75 ,85 ,95 105 . a n d o r a n d 11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216. Sol: (4, 6, 8, 10) or (10, 8, 6, 4) 12. Divide 32 into four parts which are the four terms of an AP such that the product of the first and fourth terms is to product of the second and the third terms as 7:15. Sol: Let the four parts in AP be 3 , , 3 . a d a d a d a n d a d Then, 3 3 32 4 32 8 ......... 1 a d a d a d a d a a Also, 2 2 2 2 2 2 2 2 2 3 3 : 7 :15 8 3 8 3 7 (1) 8 8 15 64 9 7 64 15 15 64 9 7 64 960 135 448 7 135 7 960 448 128 512 4 2 a d a d a d a d d d F rom d d d d d d d d d d d d d When 8 a and 2, d 3 8 3 2 8 6 2 8 2 6 8 2 10 3 8 3 2 8 6 14 a d a d a d a d When 8 a and 2, d 3 8 3 2 8 6 14 8 2 8 2 10 8 2 6 3 8 3 2 8 6 2 a d a d a d a d Hence, the four parts are 2, 6, 10 and 14. 13. The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP. Sol: Let the first three terms of the AP be , . a d a a n d a d Then,Read More

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