# The sum of two numbers is 13/6,and an even number of arithematic means are inserted between them such that their sum exceeds their number by 1.find the number of means inserted.

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given, X + Y = 13 / 6 --- -(1)

2 k is even number, where k is a natural number.

There are 2k number of Arithmetic means between X and Y.

So X, X+d, X + 2d ,.... , X+ 2k d, Y are in Arithmetic series.

X + (2k+1) d = Y

X + (2K+1) d = 13/6 - X

2 X + (2k+1) d = 13/6 --- (2)

sum of arithmetic means : X+d , X+2d,.... X+2k d

= [ X+ d + X + 2k d ] 2k / 2

= [ 2 X + (2k+1) d ] k = 2k + 1 given, as their number plus 1.

=> 13/6 * k = 2k + 1 using (2)

=> k / 6 = 1

=> k = 6

The number of means inserted are 2k = 12.

Using (2), we get , 2 X + 13 d = 13/6

d = 1/6 - 2 X /13

We can have many number of such series.. Depending on the first number, the common difference of the series can be calculated.

2 k is even number, where k is a natural number.

There are 2k number of Arithmetic means between X and Y.

So X, X+d, X + 2d ,.... , X+ 2k d, Y are in Arithmetic series.

X + (2k+1) d = Y

X + (2K+1) d = 13/6 - X

2 X + (2k+1) d = 13/6 --- (2)

sum of arithmetic means : X+d , X+2d,.... X+2k d

= [ X+ d + X + 2k d ] 2k / 2

= [ 2 X + (2k+1) d ] k = 2k + 1 given, as their number plus 1.

=> 13/6 * k = 2k + 1 using (2)

=> k / 6 = 1

=> k = 6

The number of means inserted are 2k = 12.

Using (2), we get , 2 X + 13 d = 13/6

d = 1/6 - 2 X /13

We can have many number of such series.. Depending on the first number, the common difference of the series can be calculated.