Question

If 4 AM's are inserted between 1/2 and 3 then 3rd AM IS-​

Answer 1

Answer:

The third A.M. is 2

Solution:

If we insert four Arithmetic Means between 1/2 and 3, we are making an Arithmetic Progression with six terms, of which 1/2 is the first term and 3 is the sixth term.

Let d is the common difference.

Then 6th term = 3 gives

1/2 + (6 - 1)d = 3

or, 1/2 + 5d = 3

or, 5d = 3 - 1/2

or, 5d = 5/2

or, d = 1/2

Then the third Arithmetic Mean is

= the fourth term of the Arithmetic Progression containing six terms

= 1/2 + (4 - 1) (1/2)

= 1/2 + 3/2

= (1 + 3)/2

= 4/2

= 2