Q6 of 167 Page 19

Insert A.M.s between 7 and 71 in such a way that the 5^th A.M. is 27. Find the number of A.M.s.

Let the series be 7, A₁, A₂, A₃, ........, A_n, 71

We know total terms in AP are n + 2

So 71 is the (n + 2)th term

We know,

A_n = a + (n - 1)d

So, A₆ = a + (6 - 1)d

a + 5d = 27 (5^th term)

d = 4

71 = (n + 2)th term

71 = a + (n + 2 - 1)d

71 = 7 + n(4)

n = 15

There are 15 terms in AP

Insert six A.M.s between 15 and - 13.

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.

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.

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.

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.