Question

please help me in solving this question​

Answer 1

Here it is interesting that the frequencies of the tuning forks are in AP with common difference 5!

Let the frequency of first tuning fork be,

a(1) = a

Then the frequency of the last (11th) one will be,

a(11) = a + (11 - 1) 5

a(11) = a + 10 × 5

a(11) = a + 50

Given that,

a(11) = 1.5 a(1)

a + 50 = 1.5 a

1.5a - a = 50

0.5a = 50

a(1) = 100 Hz

Thus the frequency of the first tuning fork is 100 Hz.

Then the frequency of the last tuning fork is,

a(11) = 150 Hz