Question:

Step 1: Express the first and last terms and calculate their A.M., G.M., and H.M. The given series is $100, 101, 102, \dots, n-1, n$. The first term is $a = 100$. The last term is $l = n$. We calculate the Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.) of $a$ and $l$: $$ \text{A.M.} = \frac{a+l}{2} = \frac{100+n}{2} $$ $$ \text{G.M.} = \sqrt{al} = \sqrt{100n} = 10\sqrt{n} $$ $$ \text{H.M.} = \frac{2al}{a+l} = \frac{2(100)(n)}{100+n} = \frac{200n}{100+n} $$ Step 2: Apply the condition for G.M. to be an integer and a term of the series. For G.M. to be a term of the series, it must be an integer. $$ \text{G.M.} = 10\sqrt{n} $$ For $10\sqrt{n}$ to be an integer, $\sqrt{n}$ must be an integer. This implies that $n$ must be a perfect square. Let $n = m^2$ for some positive integer $m$. Substituting $n=m^2$ into the expressions for the means: $$ \text{A.M.} = \frac{100+m^2}{2} $$ $$ \text{G.M.} = 10\sqrt{m^2} = 10m $$ $$ \text{H.M.} = \frac{200m^2}{100+m^2} $$ Step 3: Determine the valid range for 'm' based on the G.M. condition. The problem states that A.M., G.M., and H.M. are terms of the series. This means they must satisfy $100 \le \text{value} \le n$. For G.M., we have $100 \le 10m \le n$. From $10m \ge 100$: $$ m \ge 10 $$ From $10m \le n$, which is $10m \le m^2$: $$ m^2 - 10m \ge 0 $$ $$ m(m - 10) \ge 0 $$ Since $m$ must be positive (as $n=m^2$ and $n \ge 100$), we must have $m - 10 \ge 0$, which means $m \ge 10$. Combining these conditions, we conclude that $m$ must be an integer such that $m \ge 10$. Step 4: Analyze the conditions for A.M. and H.M. to be integers. For A.M. to be an integer: $$ \text{A.M.} = \frac{100+m^2}{2} $$ For $\frac{100+m^2}{2}$ to be an integer, $100+m^2$ must be an even number. Since $100$ is even, $m^2$ must also be even, which implies $m$ must be an even integer. For H.M. to be an integer: $$ \text{H.M.} = \frac{200m^2}{100+m^2} $$ For $\frac{200m^2}{100+m^2}$ to be an integer, $100+m^2$ must divide $200m^2$. Step 5: Test values of 'm' to find a suitable 'n'. We need an even integer $m \ge 10$. Let's test the smallest possible values for $m$: If $m=10$ (even and $\ge 10$): Then $n = m^2 = 10^2 = 100$. However, the series is $100, 101, \dots, n$. If $n=100$, the series is just $100$. This implies A.M. = G.M. = H.M. = 100. Let's check the calculated values: A.M. = $(100+100)/2 = 100$ G.M. = $10(10) = 100$ H.M. = $200(100)/(100+100) = 100$ All conditions are met. However, the question implies $n > 100$ by listing $100, 101, \dots, n-1, n$. If $n=100$, the terms $101, \dots, n-1$ would not exist. A typical interpretation is that $n$ must be greater than $100$ for the series to have more than one term. Let's look for the next possibility. If $m=12$ (even and $\ge 10$): Then $n = m^2 = 12^2 = 144$. A.M. = $(100+144)/2 = 244/2 = 122$ (integer) G.M. = $10(12) = 120$ (integer) H.M. = $200(144)/(100+144) = 200(144)/244 = 50(144)/61 = 7200/61$. This is not an integer. So $m=12$ is not the solution. If $m=14$ (even and $\ge 10$): Then $n = m^2 = 14^2 = 196$. A.M. = $(100+196)/2 = 296/2 = 148$ (integer) G.M. = $10(14) = 140$ (integer) H.M. = $200(196)/(100+196) = 200(196)/296 = 25(196)/37 = 4900/37$. This is not an integer. So $m=14$ is not the solution. If $m=16$ (even and $\ge 10$): Then $n = m^2 = 16^2 = 256$. A.M. = $(100+256)/2 = 356/2 = 178$ (integer) G.M. = $10(16) = 160$ (integer) H.M. = $200(256)/(100+256) = 200(256)/356 = 50(256)/89 = 12800/89$. This is not an integer. So $m=16$ is not the solution. If $m=18$ (even and $\ge 10$): Then $n = m^2 = 18^2 = 324$. A.M. = $(100+324)/2 = 424/2 = 212$ (integer) G.M. = $10(18) = 180$ (integer) H.M. = $200(324)/(100+324) = 200(324)/424 = 50(324)/106 = 25(324)/53 = 8100/53$. This is not an integer. So $m=18$ is not the solution. If $m=20$ (even and $\ge 10$): Then $n = m^2 = 20^2 = 400$. Let's verify all conditions: 1. $n=400 > 100$. This is consistent with the series description. 2. A.M. = $(100+400)/2 = 500/2 = 250$. $250$ is an integer and $100 \le 250 \le 400$. This condition is satisfied. 3. G.M. = $10(20) = 200$. $200$ is an integer and $100 \le 200 \le 400$. This condition is satisfied. 4. H.M. = $200(400)/(100+400) = 200(400)/500 = 2(400)/5 = 800/5 = 160$. $160$ is an integer and $100 \le 160 \le 400$. This condition is satisfied. Since all three means (A.M. = 250, G.M. = 200, H.M. = 160) are integers and lie within the range $[100, 400]$, $n=400$ is the correct value. The final answer is $\boxed{\text{400}}$.