32 tuning forks are arranged in order of increasing frequency such that frequency of last fork is double than first. 6 beats per second are produced by each two consecutive forks. The frequency of last fork is $n\times 186$. Find the value of n.
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Step 1: Let frequency of $I^{st}$ fork = n Step 2: Frequency of $2^{nd}$ fork = n + 6 Step 3: Simplifying the expression using exponent laws, we get: $2^{nd}$ = $2^{6+n-6}$ = $2^6$ $2^{n-6}$ = $64\times 2^{n-6}$ Step 4: Frequency of ${32}^{th}$ fork = n + 6(32 - 1) = 2n Step 5: Equating the frequency of $I^{st}$ and $2^{nd}$ forks, we get: n = $64\times 2^{n-6}$ Step 6: Simplifying the expression, we get: $2^{n-6}$ = $\frac{n}{64}$ Step 7: Substituting the value of $2^{n-6}$ in the expression for the frequency of ${32}^{th}$ fork, we get: Frequency of ${32}^{th}$ fork = $\frac{n}{64}$ $32\times 6$ + 6 = 2n Step 8: Simplifying the expression, we get: n + 186 = 2n Step 9: Solving for n, we get: n = 186 Hz Step 10: Frequency of last fork = $2n$ = $2\times 186$ = 372 Hz