Can you help me find the n th term of sequence's involving fraction's eg 3 over 4, 1 over 2, 3 over 8, 3 over 10, 1 over 4, 3 over 14. Having found the formula please help me find the 9 th term of the sequence. Can you show me some more example's of sequence's involving fraction's and explanation's of how to work out the formula and 9 the term
2006-09-20 05:20:36 · 5 answers · asked by helen y 1 in Science & Mathematics ➔ Mathematics
3/4, 1/2, 3/8. 3/10, 1/4, 3/14
3/4, 3/6, 3/8, 3/10, 3/12. 3/14
The numerator is 3 and the denominator is the multiple of two.
So the formula is 3/(2n), n>1. So when n=10 you get the ninth term. 3/(2*10) = 3/20.
2006-09-20 05:25:33 · answer #1 · answered by thierryinho 2 · 0⤊ 0⤋
Are you comfortable finding nth terms of linear sequences?
Eg 3, 5, 7, 9, ..... the nth term is 2n+1
If so, if you have a linear sequence made of up of fractions
Eg 3/5, 5/8, 7/11, 9/14
look at the numerators....its the sequence 2n+1 again.
look at the denominators...5, 8, 11, 14, ..... this is the sequence described by 3n+2.
Overall, the fractional sequence can be described using these two separate nth terms, just put the 2n+1 as the numerator and the 3n+2 as the denominator and you have your nth term
2n+1/3n+2
The 9th term is then found by substituting 9 in for the n, so in my example it would be 19/29
This always works if both sets of numbers are linear sequences.
The guy who multiplied to get all numerators to be 3 generates the sequence
3/4, 3/6, 3/8, 3/10, 3/12 and as an nth term this is
3/2n+2 and the 9th term (by substitution) is 3/20
I can't be bothered to explain quadratic sequences!
2006-09-20 05:41:04 · answer #2 · answered by Mr Glenn 5 · 0⤊ 0⤋
The given series contains simplified fractions.
The series is 3/4, 1/2, 3/8, 3/10, 1/4, 3/14
which is, 3/4, 3/6, 3/8, 3/10, 3/12, 3/14
So, the formula for the terms of this series is 3/(2n+2)
Putting n=9,
we get the ninth term as 3/20.
In any fractional series, there is no particular one formula to find the terms. You have to find the relation between two consecutive terms of the series. See if the relation is true for all consecutive terms of the series.
This relation is the formula to find the terms of the series.
2006-09-20 06:21:32 · answer #3 · answered by noesis 2 · 0⤊ 0⤋
First term is 3/4 and easy ratio is two/3 => nth term is (3/4)(2/3)^(n - a million) i.e. (3/4)(2/3)^n(2/3)?¹ => (3/2)(3/4)(2/3)^n so, (9/8)(2/3)^n Sn = (3/4)(a million - (2/3)^n)/(a million - (2/3)) => Sn = (9/4)(a million - (2/3)^n) i.e. S10 = (9/4)(a million - (2/3)¹?) so, (9/4)(58025/59049) = 2.21 (2 d.p.) :)>
2016-12-12 11:39:46 · answer #4 · answered by Erika 4 · 0⤊ 0⤋
3/4, 1/2, 3/8, 3/10, 1/4, 1/14.
Try multiplying them so the top number on each fraction is a 3:
3/4,3/6,3/8,3/10,3/12,3/14
for n=1,2,3,4,....
3/ (n+1)*2 for each term. So n=9 gives 3/(9+1)*2 = 3/20
OK?
2006-09-20 05:27:23 · answer #5 · answered by stevensontj 3 · 0⤊ 1⤋