Question

If $A_1, A_2, A_3,..... A_{51}$ are arithmetic means inserted between the numbers $a$ and $b$, then find the value of $\left(\frac{b+A_{51}}{b-A_{51}}\right) - \left(\frac{A_1+a}{A_1-a}\right)$.

• A. 102
• B. 104
• C. 51
• D. 52

Answer 1

Answer: (A)

102

Answer 2

If $A_1, A_2, \dots, A_{51}$ are $51$ arithmetic means inserted between $a$ and $b$, then the sequence $a, A_1, A_2, \dots, A_{51}, b$ forms an arithmetic progression (AP) with $53$ terms. Let $d$ be the common difference. The $k$-th term of an AP is given by $T_k = \text{first term} + (k-1)d$. The 53rd term is $b = a + (53-1)d = a + 52d$. Thus, $d = \frac{b-a}{52}$.

The first arithmetic mean, $A_1$, is the second term of the AP: $A_1 = a+d$. The 51st arithmetic mean, $A_{51}$, is the 52nd term of the AP: $A_{51} = a+51d$. Alternatively, $A_{51}$ is the second term from the end, so $A_{51} = b-d$.

We need to find the value of $E = \left(\frac{b+A_{51}}{b-A_{51}}\right) - \left(\frac{A_1+a}{A_1-a}\right)$.

Substitute $A_{51} = b-d$ into the first part: $\frac{b+(b-d)}{b-(b-d)} = \frac{2b-d}{d}$

Substitute $A_1 = a+d$ into the second part: $\frac{(a+d)+a}{(a+d)-a} = \frac{2a+d}{d}$

Now, subtract the second part from the first: $E = \frac{2b-d}{d} - \frac{2a+d}{d} = \frac{(2b-d) - (2a+d)}{d} = \frac{2b-2a-2d}{d} = \frac{2(b-a)}{d} - 2$.

Since $d = \frac{b-a}{52}$, we have $\frac{b-a}{d} = 52$. Substituting this value: $E = 2(52) - 2 = 104 - 2 = 102$.

Alternatively, there is a property for $n$ AMs inserted between $a$ and $b$: the expression $\left(\frac{b+A_n}{b-A_n}\right) - \left(\frac{A_1+a}{A_1-a}\right)$ equals $2n$. In this case, $n=51$, so the value is $2 \times 51 = 102$.