Find the sum of the sequence of odd natural numbers which are greater than 46 and less than 92.
The natural numbers are the counting numbers or positive integers. We are dealing in this question with the odd numbers that are greater than 46 and less than 92. This is the set of odd numbers 47, 49, 51, and so on, all the way up to 91. This is an arithmetic sequence, with first term 𝑎 equal to 47, common difference 𝑑 equal to two, and last term 𝑙 equal to 91. We know that the 𝑛th term 𝑎 sub 𝑛 of an arithmetic sequence is equal to 𝑎 plus 𝑛 minus one multiplied by 𝑑. We can therefore calculate the number of terms in the sequence by substituting in our values. When we substitute in these values, we get 91 is equal to 47 plus 𝑛 minus one multiplied by two.
Distributing the parentheses makes the right-hand side 47 plus two 𝑛 minus two. This simplifies to two 𝑛 plus 45. We can then subtract 45 from both sides of this equation, giving us 46 is equal to two 𝑛. Dividing both sides by two gives us 𝑛 is equal to 23. There are 23 terms in our sequence. We can now calculate the sum of the sequence by using one of the following formulae, either 𝑠 sub 𝑛 is equal to 𝑛 over two multiplied by 𝑎 plus 𝑙 or 𝑠 sub 𝑛 is equal to 𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑.
In this question, we will use the first formula. However, we would get the same answer if we use the second. The sum of our 23 terms is equal to 23 over two multiplied by 47 plus 91. This is equal to 1587, which is the sum of the sequence of odd natural numbers which are greater than 46 and less than 92.