Question

At a dance party a group of boys and girls exchange dances as follows: one boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all girls. If $b$ represents the number of boys and $g$ the number of girls, then:

Answer 1

Hint: In this question use the concept of arithmetic progression in which two consecutive terms of the sequence will always have a constant difference. Use the formula ${a_n} = a + (n - 1)d$ to find the last term.

Complete step-by-step solution -
According to the question, one boy dances with $5$ girls and second boy dances with $6$ girls and so on, we can easily see that the numbers of boys with the 6 more girls than his own number.
Hence, set up the following table, a one to one correspondence between the elements of boys set and the girls set:
Boys number $ = 1,2,3...........b$
Girls danced with $ = 5,6,7,8, 9.......g$
For each term in the sequence of boys, the term in girls is 4 more.
Hence
Boys number $ = 1,2,3........b$
Girls danced with $ = 5,6,7,8......b + 4$
$\therefore g = b + 4;b = g - 4$
Or we can use the formula for the last term of arithmetic sequence $5,6,7,8......g$
Last term ${a_n} = a + (n - 1)d$ where $a$ is the first term of the series , $n$ is the number of terms and $d$ is the common difference .
Hence, $g = 5 + (b - 1) \times (1)$$;b = g - 4$

Note: For such type of question where the increase or decrease in the series, there we use arithmetic progression. From the above question we can understand that there are $b$ number of boys (means the ${n^{th}}$ term of boys is $b$) and $g$ number of girls( means the ${n^{th}}$term of girls is $g$) and when one boy dance then $5$ girls dance with him , when second boy dance then $6$ girls dance with him and so on till there ${n^{th}}$term , So we can say that the boys and girls are in arithmetic sequence