If 13 arithmetic means are inserted between 1 and \(\rm\frac{9}{2} \), then find the 4^th arithmetic mean.

CONCEPT:

• If a₁, a₂, ……, a_n are n numbers, then the arithmetic mean of these numbers can be expressed as \(A.M = \rm\frac{(a_1+a_2+.....a_n )}{n}\).

• The n numbers A₁, A₂, ……, A_n are said to be A.M.’s between the numbers p and q if p, A₁, A₂, ……, A_n, q are in A.P, if  d is the common difference of this A.P., then q = a + (n + 2 – 1) d where \(\rm d = \rm\frac{q-p}{n + 1} \)

CALCULATION :

Let a₁, a₂, a₃, a₄ . . . . . . .a₁₃ be 13 arithmetic means between 1 and 9/2

⇒ 1, a₁, a₂, ……, a_13, \(\frac{9}{2}\) will be in A.P.

As we know that, the n numbers A1, A2, ……, An are said to be A.M.’s between the numbers p and q if p, A1, A2, ……, An, q are in A.P, if  d is the common difference of this A.P., then q = a + (n + 2 – 1) d where \(\rm d = \rm\frac{q-p}{n + 1} \)

Now \(\frac{9}{2}=1 + 14d\)

On solving we get  \(d=\frac{1}{4}\)

Hence \(a\rm_{4} = 1 + 4\rm\frac{(1)}{(4)}= 2\)

Therefore, the fourth arithmetic mean is 2.

Hence, option(2)is the correct answer.