Answers: (i) A be the Arithmetic mean 7, A, 13 are in AP A-7 = 13-A 2A = 13 + 7 A = 10 (ii) A be the Arithmetic mean 12, A, – 8 are in AP A – 12 = – 8 – A 2A = 12 + 8 A =...
Find the A.M. between: (i) 7 and 13 (ii) 12 and – 8
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Exercise 19.6
Find the A.M. between: (x – y) and (x + y)
Answer: A be the Arithmetic mean x – y, A, x + y are in AP A – (x – y) = (x + y) – A 2A = x + y + x – y A = x
Insert 4 A.M.s between 4 and 19.
Answer: A1, A2, A3, A4 - 4 AM Between 4 and 19 4, A1, A2, A3, A4, 19 are in AP. By using the formula, d = (b-a) / (n+1) d = (19 – 4) / (4 + 1) d = 15/5 d = 3 A1 = a + d = 4 + 3 = 7 A2 = A1 + d = 7 +...
Insert 7 A.M.s between 2 and 17.
Answer: A1, A2, A3, A4, A5, A6, A7 - 7 AMs between 2 and 17 2, A1, A2, A3, A4, A5, A6, A7, 17 are in AP By using the formula, an = a + (n – 1)d an = 17, a = 2, n = 9 17 = 2 + (9 – 1)d 17 = 2 + 9d –...
Insert six A.M.s between 15 and – 13.
Answer: A1, A2, A3, A4, A5, A6 - 7 AM between 15 and – 13 15, A1, A2, A3, A4, A5, A6, – 13 are in AP an = a + (n – 1)d an = -13, a = 15, n = 8 -13 = 15 + (8 – 1)d -13 = 15 + 7d 7d = -13 – 15 7d =...
There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n
Answer: Let the series be 3, A1, A2, A3, …….., An, 17 Given, an/a1 = 3/1 Total terms in AP are n + 2 17 is the (n + 2)th term By using the formula, An = a + (n – 1)d An = 17, a = 3 So, 17 = 3 + (n +...
Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.s.
Answer: Let the series be 7, A1, A2, A3, …….., An, 71 Total terms in AP are n + 2 71 is the (n + 2)th term By using the formula, An = a + (n – 1)d An = 71, n = 6 A6 = a + (6 – 1)d a + 5d = 27...
If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
Answer: Let a and b be the first and last terms The series be a, A1, A2, A3, …….., An, b Mean = (a+b)/2 Mean of A1 and An = (A1 + An)/2 A1 = a+d An = a – d AM = (a+d+b-d)/2 AM = (a+b)/2 AM between...
If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.
Answer: Given, A1 = AM of x and y A2 = AM of y and z A1 = (x+y)/2 A2 = (y+x)/2 AM of A1 and A2 = (A1 + A2)/2 => [(x+y)/2 + (y+z)/2]/2 => [x+y+y+z]/2 => [x+2y+z]/2 x, y, z are in AP, y =...
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P
Answer: A1, A2, A3, A4, A5 - 5 numbers between 8 and 26 8, A1, A2, A3, A4, A5, 26 are in AP By using the formula, An = a + (n – 1)d An = 26, a = 8, n = 7 26 = 8 + (7 – 1)d 26 = 8 + 6d 6d = 26 – 8 6d...