UPSC CISF-AC-EXE

Now we need to find the value of $a-b$. We know the identity $(a-b)^2 = a^2 – 2ab + b^2$.
We can rewrite this as $(a-b)^2 = (a^2 + b^2) – 2ab$.
Substitute the values of $a^2+b^2$ and $ab$ that we found:
$(a-b)^2 = 52 – 2(24)$
$(a-b)^2 = 52 – 48$
$(a-b)^2 = 4$.
Taking the square root of both sides:
$a-b = \pm \sqrt{4} = \pm 2$.
The problem states that $a > b$, which means $a-b$ must be a positive value.
Therefore, $a-b = 2$.

Hold on, let’s re-calculate the subtraction step correctly.
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$2S – S = (1\cdot2^2 – 2\cdot2^2) + (2\cdot2^3 – 3\cdot2^3) + \dots + (9\cdot2^{10} – 10\cdot2^{10}) + 10\cdot2^{11} – 1\cdot2^1$
This is incorrect. The terms should align:
$S = \quad 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
Subtracting S from 2S:
$S = (10\cdot2^{11}) – (1\cdot2^1) – [(2-1)2^2 + (3-2)2^3 + \dots + (10-9)2^{10}]$
$S = 10\cdot2^{11} – 2 – [2^2 + 2^3 + \dots + 2^{10}]$
The geometric series is $2^2 + 2^3 + \dots + 2^{10}$. First term $a=2^2=4$, ratio $r=2$, number of terms $n=9$.
Sum $G = 4 \frac{2^9 – 1}{2-1} = 4(512 – 1) = 4 \times 511 = 2044$.
Let’s re-calculate $G$ using the formula $a(r^n-1)/(r-1)$ where $a=2$, $n=10$ for $2^1 + … + 2^{10}$, then subtract the first term $2^1$.
$G’ = 2^1 + 2^2 + \dots + 2^{10} = 2 \frac{2^{10} – 1}{2-1} = 2(1024 – 1) = 2 \times 1023 = 2046$.
The sum $2^2 + 2^3 + \dots + 2^{10} = G’ – 2^1 = 2046 – 2 = 2044$.
$S = 10\cdot2^{11} – 2 – 2044 = 10\cdot2^{11} – 2046$.
$10 \cdot 2^{11} = 10 \cdot 2048 = 20480$.
$S = 20480 – 2046 = 18434$.

Now check the options:
A) $9\times2^{11} + 2 = 9 \times 2048 + 2 = 18432 + 2 = 18434$. Matches.
B) $10\times2^{11} + 2 = 10 \times 2048 + 2 = 20480 + 2 = 20482$.
C) $10\times2^{11} – 2 = 10 \times 2048 – 2 = 20480 – 2 = 20478$.
D) $9\times2^{11} – 2 = 9 \times 2048 – 2 = 18432 – 2 = 18430$.

My original calculation of the geometric series part in the subtraction was correct:
$S = 10 \cdot 2^{11} – (2^1 + 2^2 + \dots + 2^{10})$
The geometric series $2^1 + 2^2 + \dots + 2^{10}$ has $a=2, r=2, n=10$.
Sum is $2 \frac{2^{10}-1}{2-1} = 2(1024-1) = 2(1023) = 2046$.
$S = 10 \cdot 2^{11} – 2046$.
$S = 10 \cdot 2048 – 2046 = 20480 – 2046 = 18434$.
This matches $9 \cdot 2^{11} + 2$.

Let’s re-do the AGP subtraction step structure:
$S = 1\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
Subtracting the first from the second, aligning terms vertically:
$S = (2\cdot2^2 – 1\cdot2^2) + (3\cdot2^3 – 2\cdot2^3) + \dots + (10\cdot2^{10} – 9\cdot2^{10}) + 10\cdot2^{11} – 1\cdot2^1$ — This is wrong alignment
Correct alignment for subtraction:
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$

$2S – S = (10\cdot2^{11}) + (9\cdot2^{10} – 10\cdot2^{10}) + \dots + (2\cdot2^2 – 3\cdot2^2) + (1\cdot2^1 – 2\cdot2^1)$ — Also wrong alignment.

Correct method:
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$

$2S – S = (10\cdot2^{11}) – (1\cdot2^1) + (1\cdot2^2 – 2\cdot2^2) + (2\cdot2^3 – 3\cdot2^3) + \dots + (9\cdot2^{10} – 10\cdot2^{10})$ This is also not right.

The general formula for $\sum_{k=1}^n k r^k$ is $\frac{nr^{n+2} – (n+1)r^{n+1} + r}{(r-1)^2}$.
Here $n=10$, $r=2$.
$S = \frac{10 \cdot 2^{12} – (10+1)2^{11} + 2}{(2-1)^2} = \frac{10 \cdot 2^{12} – 11 \cdot 2^{11} + 2}{1^2}$
$S = 10 \cdot 2 \cdot 2^{11} – 11 \cdot 2^{11} + 2$
$S = 20 \cdot 2^{11} – 11 \cdot 2^{11} + 2$
$S = (20 – 11) \cdot 2^{11} + 2$
$S = 9 \cdot 2^{11} + 2$.
This matches option A and confirms the previous calculation result.

The error in manual subtraction breakdown was in the terms. It should be:
$S = 1\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$2S – S = (10\cdot2^{11}) + (2\cdot2^2 – 1\cdot2^2) + (3\cdot2^3 – 2\cdot2^3) + \dots + (10\cdot2^{10} – 9\cdot2^{10}) – 1\cdot2^1$
This is still not quite right. The terms are offset.
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + (n-1)r^{n-1} + n r^n$
$rS = \quad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1)r^n + n r^{n+1}$
$rS – S = n r^{n+1} – 1\cdot r^1 – [(2-1)r^2 + (3-2)r^3 + \dots + (n-(n-1))r^n]$
$(r-1)S = n r^{n+1} – r – [r^2 + r^3 + \dots + r^n]$
The geometric series in the bracket is $r^2 + \dots + r^n$. First term $r^2$, ratio $r$, number of terms $n-1$.
Sum is $r^2 \frac{r^{n-1}-1}{r-1}$.
$(r-1)S = n r^{n+1} – r – r^2 \frac{r^{n-1}-1}{r-1}$. For $r=2$:
$S = n 2^{n+1} – 2 – 4 \frac{2^{n-1}-1}{1} = n 2^{n+1} – 2 – 4(2^{n-1}-1) = n 2^{n+1} – 2 – 2^2 2^{n-1} + 4 = n 2^{n+1} – 2^{n+1} + 2$.
$S = (n-1)2^{n+1} + 2$.
With $n=10$:
$S = (10-1)2^{10+1} + 2 = 9\cdot2^{11} + 2$.
This confirms the formula derived from the difference method.

Let’s verify with the other examples:
DOG -> FQI
D(4) -> F(6) : +2
O(15) -> Q(17) : +2
G(7) -> I(9) : +2 (Consistent)

SUN -> UWP
S(19) -> U(21) : +2
U(21) -> W(23) : +2
N(14) -> P(16) : +2 (Consistent)

Applying the +2 rule to BET:
B(2) -> D(4)
E(5) -> G(7)
T(20) -> V(22)
The resulting letters are D, G, and V.
So, the code word for BET is DGV.

Alternatively, assume the first walk of 1 km is along the positive y-axis. The position is (0,1).
Turning right from the positive y-axis means moving along the positive x-axis. The walk of 2 km leads to the position (2, 1).
Turning left from the positive x-axis means moving along the positive y-axis. The walk of 3 km leads to the position (2, 1+3) = (2, 4).
Point Q is at (2, 4). The distance from P(0,0) to Q(2,4) is $\sqrt{(2-0)^2 + (4-0)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
In both common interpretations of direction and turns, the distance is $2\sqrt{5}$ km.