Sequences - nth Terms and Limiting Values

Work out what is added each time to get from one term to the next.

1. Each term is x-4 more than the one before
2. 4th term =3x+1+x-4=4x-3
3. 5th term =4x-3+x-4=5x-7
4. Answer: 5x-7

When n is very large, the numbers 5 and 3 hardly matter: compare the terms in n on the top and the bottom.

1. For large n, the nth term is close to \dfrac{-8n}{4n}
2. \dfrac{-8n}{4n}=-2
3. Answer: -2
4. (\frac53 uses the constants instead of the terms in n; 2 loses the minus sign; -8 ignores the bottom.)

The common difference goes in front of n; then work out what you must add to make the 1st term correct.

1. Common difference =-11-(-17)=6, so the nth term starts 6n
2. When n=1, 6n=6, but the 1st term is -17
3. -17-6=-23, so subtract 23
4. Check: n=4 gives 24-23=1 ✓
5. Answer: 6n-23

n is the position of a term, so it must be a positive whole number.

1. If 150 were a term, then 7n+2=150 for some position n
2. 7n=148, so n=\dfrac{148}{7}=21.14\ldots
3. A position must be a whole number, so 150 is not a term
4. Answer: 7n+2=150 gives n=\frac{148}{7}, which is not a whole number
5. (The 2nd term is 16, so not every term is odd; the 1st term is 9, so terms need not be multiples of 7; n is positive here.)

Going from the 1st term to the 10th term adds the common difference 9 times, not 10.

1. From the 1st term to the 10th term is 9 steps
2. Common difference =27\div 9=3
3. So the nth term starts 3n
4. When n=1, 3n=3, but the 1st term is 4.5, so add 1.5
5. Check: 10th term =30+1.5=31.5, which is 4.5+27 ✓
6. Answer: 3n+1.5

Half of the second difference is the coefficient of n^2.

1. First differences: 9,\ 15,\ 21
2. Second difference: 6, so a=6\div2=3
3. 3n^2 gives 3,\ 12,\ 27,\ 48
4. Each term is 4 more than 3n^2, so b=4
5. Answer: 3n^2+4

Put the nth term of sequence B in a bracket before you subtract it.

1. nth term of C =(4n+7)-(10-3n)
2. =4n+7-10+3n
3. Check: 1st terms are 11 and 7, and 11-7=4=7-3 ✓
4. Answer: 7n-3

Think about the n^2 term you would get if you expanded the top.

1. Expanding the top gives 2n^2-5n-3
2. For large n, the n^2 terms are much bigger than the rest
3. So the nth term is close to \dfrac{2n^2}{4n^2}=\dfrac12
4. Answer: \dfrac12

Set the nth term equal to 3 and multiply both sides by (2n+3).

1. \dfrac{7n-2}{2n+3}=3
2. Multiply by (2n+3): 7n-2=6n+9
3. n=11
4. Check: \dfrac{77-2}{22+3}=\dfrac{75}{25}=3 ✓
5. Answer: the 11th term

Substitute n=9 and n=3 separately, simplify each fraction, then subtract.

1. 9th term =\dfrac{45}{15}=3
2. 3rd term =\dfrac{15}{9}=\dfrac53
3. Difference =3-\dfrac53
4. Answer: \dfrac43

Form a quadratic equation equal to zero and factorise it; remember that n is a position.

1. n^2-6n+20=92
2. n^2-6n-72=0
3. (n-12)(n+6)=0, so n=12 or n=-6
4. n is a position, so it must be a positive whole number
5. Answer: the 12th term

Find the second difference and halve it to get the coefficient of n^2.

1. First differences: -1, \ -5, \ -9
2. Second difference: -4, so the n^2 term is -2n^2
3. -2n^2 gives -2, \ -8, \ -18, \ -32
4. Sequence minus -2n^2: 6, \ 11, \ 16, \ 21, which is 5n+1
5. Check: n=3 gives -18+15+1=-2 ✓
6. Answer: -2n^2+5n+1

The second difference is odd, so the coefficient of n^2 is a fraction.

1. First differences: 4,\ 7,\ 10
2. Second difference: 3, so the n^2 term is 1.5n^2
3. 1.5n^2 gives 1.5,\ 6,\ 13.5,\ 24
4. Sequence minus 1.5n^2: 0.5,\ 0,\ -0.5,\ -1
5. That is -0.5n+1
6. Answer: 1.5n^2-0.5n+1

Write down the nth term of each sequence, then solve an inequality.

1. Sequence A: 4n+4
2. Sequence B: difference -2.5, so 302.5-2.5n
3. 4n+4>302.5-2.5n
4. 6.5n>298.5, so n>45.92\ldots
5. The first whole number is n=46
6. Check: 45th terms 184<190; 46th terms 188>187.5 ✓
7. Answer: 46

The 25th term is the 1st term plus 24 lots of the common difference.

1. Common difference =3k-2
2. 25th term =2k+1+24(3k-2)
3. =74k-47
4. 74k-47=175, so 74k=222
5. Answer: k=3

Expand both brackets before subtracting, and put the second expansion in a bracket.

Part (a)

1. (n+5)(n+4)=n^2+9n+20
2. (n+2)(n+1)=n^2+3n+2
3. Subtract: n^2+9n+20-(n^2+3n+2)=6n+18
4. Factorise: 6n+18=6(n+3)
5. Answer: 6(n+3)

Part (b)

1. P - Q =6(n+3) and n+3 is an integer, so P - Q is always a multiple of 6
2. When n=2, P - Q =30, which is not a multiple of 12, 9 or 18
3. Answer: P - Q is a multiple of 6

Part (c)

1. (n+5)(n+4)=342
2. n^2+9n+20=342, so n^2+9n-322=0
3. (n-14)(n+23)=0, so n=14 or n=-23
4. n must be a positive whole number
5. Answer: the 14th term

Use 6n+1 with n=1,2,3 to write down the first few terms. Answer part (a) before attempting part (b).

Part (a)

1. Put n=1,\ 2,\ 3 into 6n+1: the first differences are 7,\ 13,\ 19
2. So the sequence starts 4,\ 11,\ 24,\ 43
3. Second difference 6, so a=3
4. Sequence minus 3n^2: 1,\ -1,\ -3,\ -5, which is -2n+3
5. Answer: 3n^2-2n+3

Part (b)

1. 3n^2-2n+3=1163
2. 3n^2-2n-1160=0
3. (3n+58)(n-20)=0, so n=20 or n=-\frac{58}{3}
4. n must be a positive whole number
5. Answer: the 20th term

For (c), write the limiting value minus the nth term as a single fraction.

Part (a)

1. For large n, the nth term is close to \dfrac{5n^2}{n^2}
2. Answer: 5

Part (b)

1. \dfrac{5n^2-2}{n^2+4}=4.45
2. 5n^2-2=4.45n^2+17.8
3. 0.55n^2=19.8, so n^2=36
4. n=6 (n=-6 is not a position)
5. Answer: the 6th term

Part (c)

1. Difference =5-\dfrac{5n^2-2}{n^2+4}=\dfrac{22}{n^2+4}
2. \dfrac{22}{n^2+4}<0.01, so n^2+4>2200
3. n^2>2196, so n>46.86\ldots
4. Check: 46^2=2116 is too small; 47^2=2209 works
5. Answer: n=47

For (b), set the two nth terms equal and clear the fraction; remember n must be a positive whole number.

Part (a)

1. For large n, the 21 and the 3 hardly matter
2. \dfrac{4n+21}{n+3}\approx\dfrac{4n}{n}=4
3. Answer: 4

Part (b)

1. Set the nth terms equal: \dfrac{4n+21}{n+3}=2n-7
2. Multiply by (n+3): 4n+21=(2n-7)(n+3)=2n^2-n-21
3. Rearrange: 2n^2-5n-42=0
4. Factorise: (2n+7)(n-6)=0, so n=-3.5 or n=6
5. n is a position, so it must be a positive whole number
6. Answer: n=6 (both terms equal 5)

Part (c)

1. Solve \dfrac{4n+21}{n+3}>4.2
2. n+3 is positive, so 4n+21>4.2n+12.6
3. 8.4>0.2n, so n<42
4. The 42nd term equals exactly 4.2, which is not greater than 4.2
5. The terms decrease towards 4, so terms 1 to 41 are greater than 4.2
6. Answer: 41

Find the nth term of A and of B first, then multiply the two brackets. Answer part (a) before attempting part (b).

Part (a)

1. Sequence A: 2n+1
2. Sequence B: 11-n
3. (2n+1)(11-n)=22n-2n^2+11-n
4. Answer: -2n^2+21n+11

Part (b)

1. The turning point of -2n^2+21n+11 is at n=\dfrac{21}{4}=5.25
2. So the terms are greatest when n is closest to 5.25
3. 5th term =11\times6=66; 6th term =13\times5=65
4. Answer: 66

Part (c)

1. A term is positive when (2n+1)(11-n)>0
2. 2n+1 is always positive, so we need 11-n>0, which is n<11
3. The 11th term is 23\times0=0, which is not positive
4. Answer: 10 terms