Special Sequences - Higher

Identify the whole numbers that have been cubed.

1. These terms are 2^3,3^3,4^3,5^3.
2. The next term is 6^3=216.

The amounts added increase by one each time.

1. The increases are 2,3,4,5.
2. The next increase is 6, giving 15+6=21.

Use the last two displayed terms.

1. Add the previous two terms: 11+18=29.

Divide one term by the term before it to find the common multiplier.

1. Each term is multiplied by \frac12.
2. The next term is 1.5\times\frac12=0.75, or \frac34.

A geometric sequence has the same multiplier between consecutive terms.

1. The terms 3,6,12,24 each come from multiplying the previous term by 2.
2. The other options have changing ratios, so only the third sequence is geometric.

The first term needs no multiplication by the common ratio.

1. The first term is 4 and each term is 3 times the one before, so the common ratio is 3.
2. To reach term n, multiply the first term by 3 a total of n-1 times: 4\times3^{n-1}.
3. Check: n=1 gives 4\times3^0=4. The option 4\times3^n gives 12 for the first term, so it is wrong.

Divide the second term by the first, including the signs.

1. The common ratio is -6\div18=-\frac13.
2. Check: multiplying -6 by -\frac13 gives the third term, 2.

Work backwards using subtraction instead of addition.

1. The third term equals the sum of the first two, so a+7=12.
2. Subtract 7 to get a=5.

Keep the negative sign in the common ratio as you continue the sequence.

1. The common ratio is -2.
2. The fifth term is 48, so the sixth term is 48\times(-2)=-96.

Find the two square numbers from their positions before subtracting.

1. The twelfth square number is 12^2=144 and the eleventh is 11^2=121.
2. The difference is 144-121=23.

Set the formula equal to the given value and look for consecutive factors of twice that value.

1. Set the formula equal to 276: \frac{n(n+1)}{2}=276, so n(n+1)=552.
2. n and n+1 are consecutive whole numbers with product 552. Since 23^2=529 and 24^2=576, try 23\times24=552.
3. So n=23: 276 is the 23rd triangular number.

Express the third and fourth terms using an unknown first term.

1. Let the first term be a. The third term is a+8.
2. The fourth term is 8+(a+8)=a+16.
3. So a+16=21, giving a=5.

Divide out the first-term factor and recognise the resulting power.

1. The equation is 3\times2^{n-1}=192, so 2^{n-1}=64.
2. Since 64=2^6, n-1=6.
3. Therefore n=7.

Continue dividing by three, keeping track of the position numbers.

1. The first five terms are 162,54,18,6,2.
2. The sixth term is \frac23, which is below 1. All previous terms are above 1.
3. The required position is 6.

Find the common ratio, then apply it twice to the third term.

1. The ratio is \frac{\sqrt6}{2}. Its square is \frac64=\frac32.
2. The fifth term is two steps after the third, so multiply the third term by the square of the ratio.
3. The fifth term is 3\times\frac32=\frac92.

Find the ratio from the given terms, then divide to move backwards.

1. The common ratio is \frac{3\sqrt2}{6}=\frac{\sqrt2}{2}.
2. The first term is 6\div\frac{\sqrt2}{2}=\frac{12}{\sqrt2}.
3. Rationalising and simplifying gives 6\sqrt2.

Write the later terms in terms of the two starting values before solving the equations.

Part (a)

1. Write the terms using a and b: the third term is a+b and the fourth is b+(a+b)=a+2b.
2. The fifth term is (a+b)+(a+2b)=2a+3b.
3. So a+b=11 and 2a+3b=27.
4. Multiply the first equation by 3: 3a+3b=33. Subtract the second equation: a=6.

Part (b)

1. Use a+b=11 with a=6: b=5.
2. Check: the sequence is 6,5,11,16,27.

Multiply out 8T_n+1 first, then look for a perfect square.

1. 8T_n=8\times\frac{n(n+1)}{2}=4n(n+1).
2. So 8T_n+1=4n^2+4n+1.
3. Compare with (an+b)^2=a^2n^2+2abn+b^2: a^2=4 and b^2=1 give a=2 and b=1, and then 2ab=4 matches the middle term.
4. So 8T_n+1=(2n+1)^2.

Equate the two consecutive ratios, then use the positivity condition.

1. Equal ratios give \frac6{2x}=\frac{3x+3}{6}.
2. Cross-multiply: 36=2x(3x+3), so x^2+x-6=0.
3. Factorise: (x+3)(x-2)=0.
4. The condition x>0 rejects -3, leaving x=2. The terms are then 4,6,9, with common ratio \frac32.

Use the arithmetic sequence to write q in terms of p, then use the equal ratios of the geometric sequence.

Part (a)

1. In an arithmetic sequence the difference is constant: p-4=q-p, so q=2p-4.
2. In a geometric sequence the ratio is constant: \frac{q}{p}=\frac{18}{q}, so q^2=18p.
3. Substitute q=2p-4: (2p-4)^2=18p, so 4p^2-16p+16=18p.
4. Rearrange and divide by 2: 2p^2-17p+8=0, which factorises as (2p-1)(p-8)=0.
5. p=\frac12 would make q=2\times\frac12-4=-3, which is negative, so p=8.

Part (b)

1. From part (a), q=2p-4 with p=8, so q=12.
2. Check: 4,8,12 go up by 4 each time, and 8,12,18 are each multiplied by \frac32.