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| Math Central | Quandaries & Queries |
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Question from Arjun, a student: 1/2 3/5 5/8 7/11- need to find the nth term. I did search the data base & found one for fractions but what I want to know is when calculating nth term for the denominator in the example give in your database how do we get (n-1)? I thank you for your support. Arjun |
Arjun,
I think you are looking at the question that Zarinah sent us. She sent the sequence
1/4, 2/7, 3/10, 4/13, 5/16, ...
In my response I let N designate numerator and D denominator and constructed the table.
| 1st term | 2nd term | 3rd term | 4th term | 5th | |
|---|---|---|---|---|---|
| N | 1 | 1 + 1 | 1 +2(1) | 1 + 3(1) | 1 + 4(1) |
| D | 4 | 4 + 3 | 4 + 2(3) | 4 + 3(3) | 4 + 4(3) |
I you look at the terms in the sequence you will see that the numerators are increasing by 1 (1, 2, 3, 4, ...) and the denominators are increasing by 3 (4, 7, 10, 13, ...). I highlighted this in the table by using the colour red. I am going to draw the table again and this time highlight a different number in blue.
| 1st term | 2nd term | 3rd term | 4th term | 5th | |
|---|---|---|---|---|---|
| N | 1 | 1 + 1(1) | 1 +2(1) | 1 + 3(1) | 1 + 4(1) |
| D | 4 | 4 + 1(3) | 4 + 2(3) | 4 + 3(3) | 4 + 4(3) |
So the sequence starts at 1/4 and then to obtain subsequent terms you add ones to the numerator and threes to the numerator. The blue numbers tell you how many times you have added 1 to the numerator and 3 to the denominator.
To obtain the second term you add 1 to the numerator and 3 to the denominator once.
To obtain the third term you add 1 to the numerator and 3 to the denominator twice.
To obtain the fourth term you add 1 to the numerator and 3 to the denominator three times.
To obtain the fifth term you add 1 to the numerator and 3 to the denominator four times.
So the number of times you add 1 to the numerator and 3 to the dominator is one less than the term number. That is
To obtain the nth term you add 1 to the numerator and 3 to the denominator n-1 times.
Thus the nth term is [1 + (n-1)(1)]/[4 + (n-1)(3)].
I hope this helps,
Penny
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