Question

Let S= {x:x is a positive multiple of 3 less than 100
P= {x:x is a prime number less than 20}; then find n(s) + n (p)

Answer 1

Hint: In this question, first we will need to find ‘S’ which is the set of all positive multiples of 3 less than 100. After that we will find ‘P’ which is the set of all prime numbers less than 20. Here we will use the concept of A.P to find the number of terms.

Complete step-by-step answer:
An arithmetic progression is a sequence of numbers such that the difference between two consecutive terms is constant. For example, the sequence 5, 7, 9, 11, 13, …. Is an arithmetic progression with a common difference of 2. And the initial term of arithmetic progression is ‘ $ {a_1} $ ’ and the common difference of successive numbers is denoted by ‘d’, then the nth term of the sequence $ {a_n} $ is given by: $ {a_n} = {a_1} + (n - 1)d $
 S= {x:x is a positive multiple of 3 less than 100}
 So, S = {3, 6, 9, 12, …………., 96, 99}
 Since, 3, 6, 9, 12, ………., 99 is an A.P.
  Where the first term, a = 3
  & Common difference, d = 3
  Let n be the number of terms,
 $ {a_n} = {a_1} + (n - 1)d $ [ where $ {a_n} $ = the nth term in the sequence or the last term, $ {a_1} $ = the first term in the sequence, d= the common difference between terms]
  99 = 3 + (n-1) 3
  $ \Rightarrow $ 3 + 3n – 3 = 99
  $ \Rightarrow $ 3n = 99
   $ \therefore $ n = $ \dfrac{{99}}{3} $ = 33
Thus, number of elements in set S, n(s) = 33
Now, P = {x:x is a prime number less than 20}
So, P = {2, 3, 5, 7, 11, 13, 17, 19}
Thus, number of elements in set P, n(p) = 8,
Hence, n(s) + n(p) = 33 + 8 = 41

Note: Prime number is that natural number which has exactly two factors, i.e., 1 and the number itself is called prime number. The lowest prime number is 2 and 2 is also the only even prime number. So, students, I hope you got the concept of prime number and also arithmetic progression (A.P.).