Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for the nth term of a quadratic sequence in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the The nth term of a quadratic sequence topic page.
When you see a sequence whose differences are not constant, do this.
Subtract each term from the next. Those gaps are the first differences.
If the first differences themselves go up by a fixed amount, that amount is the second difference.
The coefficient of n^2 is half the second difference. A second difference of 2 means n^2. A second difference of 4 means 2n^2.
Subtract the n^2 part from the sequence and find the linear term that is left.
Put n=3 into your rule. It must give the third term you started with.
If a term equals a number, set the nth term equal to it and solve. Reject a negative position.
The sequence 2, 6, 12, 20, 30 has first differences 4, 6, 8, 10 and second difference 2. Half of 2 is 1, so a=1. The square numbers n^2 are 1, 4, 9, 16, 25. Subtract them from the sequence: 2-1=1, 6-4=2, 12-9=3, 20-16=4, 30-25=5. The remainder is 1, 2, 3, 4, 5, which is just n. So the nth term is n^2+n. Check n=4: 16+4=20. Check n=3: 9+3=12.
Using the second difference itself as a is the error that doubles every term. If you had taken a=2, the squares 2n^2 would be 2, 8, 18, 32, 50, already larger than several terms of the sequence, and the remainder would go negative for no good reason. Halve first, then subtract. Write the second-difference row on the page so the halving is visible.
Method: 1. Write the first differences under the terms. 2. Write the second differences under those. 3. Halve a constant second difference to get a. 4. Do not use the second difference itself as a.
The sequence 3, 9, 19, 33, 51 has first differences 6, 10, 14, 18 and second difference 4. So a=2. Then 2n^2 is 2, 8, 18, 32, 50. Subtract from each term: 3-2=1, 9-8=1, 19-18=1, 33-32=1, 51-50=1. The remainder is the constant sequence 1, whose nth term is 1. The quadratic formula is 2n^2+1. Check the fifth term: 2(25)+1=51. Check the first: 2+1=3.
Subtract for every term, not only the first. A single subtraction can look plausible and still hide a linear piece. If the remainder is 3, 4, 5, 6, 7, that is n+2, not a constant. You only see the pattern if the whole row is written. Keep the signs. A remainder of -1, -1, -1 is the constant -1, and the formula contains -1, not a missing constant.
Method: 1. List an^2 for each position you were given, starting at n=1. 2. Subtract that list from the sequence, term by term. 3. Find the nth term of the remainder. It is linear or constant. 4. Add it to an^2 and check two positions.
Take 5, 12, 23, 38, 57. First differences 7, 11, 15, 19. Second difference 4, so a=2. Then 2n^2 is 2, 8, 18, 32, 50. Subtract: 5-2=3, 12-8=4, 23-18=5, 38-32=6, 57-50=7. The remainder 3, 4, 5, 6, 7 has nth term n+2. Therefore the sequence is 2n^2+n+2. Check n=4: 32+4+2=38. Check n=6, which you did not use: 72+6+2=80. The sixth term really is 80.
If a remainder decreases, the linear coefficient is negative. Write each subtraction as a number line if you need to: 1-2=-1, not 1. A sign error in one place shifts b and c together, and the first term may still match by accident. That is why the second check should be a later term. If term 1 matches and term 4 does not, b or c is wrong. If neither matches, a is wrong.
Method: 1. A changing remainder still has an nth term. 2. Include a negative b if the remainder falls. 3. The constant is the value that makes n=1 work, after a and b are fixed. 4. Check a later term you did not use to build the row.
To test whether 80 is in 2n^2+n+2, solve 2n^2+n+2=80, so 2n^2+n-78=0. You do not have to use the formula if you can see n=6 from the check above: 2(36)+6+2=80. A positive integer n means yes, and you should state the position. Here 80 is the 6th term. A solution such as n=2.5 or n=-3 does not count. The position has to be a positive whole number.
The 20th term of the same sequence is 2(400)+20+2=822. That is a substitution, not a new method. Do not find 20 second differences to reach it. Once the formula is checked, any position is one calculation. Say "the 20th term is 822" rather than only writing 822, so the position and the value are both on the page.
Method: 1. Set an^2+bn+c equal to the number you were given. 2. Solve for n. 3. Keep the solution only if it is a positive integer. 4. State the position, not only "yes".
The second difference can be negative or odd, and the rule is the same: the coefficient a is half of the second difference. For the sequence 8, 7, 4, -1, -8 the first differences are -1, -3, -5, -7 and the second difference is -2, so a=-1. The n^2 term is then -n^2, which gives -1, -4, -9, -16, -25. Writing the rows underneath each other in a neat table makes it far easier to spot a slip with a minus sign.
A second difference of 1 gives a=tfrac12, and the terms may be decimals such as 1.5, 4, 7.5. Do not panic when you see halves. Subtract the tfrac12n^2 values from the terms and look at what is left. In every case, write the an^2 row, subtract it, and keep the minus signs in the remainder row.
Method: 1. Work out the first differences, then the second differences. 2. Halve the second difference to find a, keeping its sign. 3. Write the row of an^2 values and subtract it from the terms. 4. Find the linear rule for what is left, using minus signs where it decreases.
Tip: A negative second difference means the sequence turns downwards in the end, so a is negative.
There is a second route that uses equations. The nth term is an^2+bn+c. Writing the first three terms gives a+b+c, 4a+2b+c and 9a+3b+c. Subtracting pairs gives 3a+b for the first difference and 5a+b for the next, so the second difference is 2a. This explains why you halve the second difference.
The same idea solves exam questions that give only two or three terms. If the nth term is an^2+bn and the second term is 6 and the third term is 15, then 4a+2b=6 and 9a+3b=15. Divide the first by 2 and the second by 3 to get 2a+b=3 and 3a+b=5. Subtracting gives a=2, and then b=-1.
Method: 1. Write the nth term as an^2+bn+c, or an^2+bn if there is no constant. 2. Substitute the given positions to make equations in a, b and c. 3. Simplify and solve by elimination. 4. Check the formula against another term in the sequence.
A question may give the nth term and ask for a particular term, such as the 12th, by substituting n=12. Be careful with order: square first, then multiply by the coefficient, then add or subtract. For 4n^2-3n at n=12 the working is 4× 144-36=540. Check each answer by thinking about whether it is sensible for the shape of the sequence.
To test whether a number is in the sequence, set the formula equal to it. For a quadratic you can often avoid solving: find the terms either side of the number. If n=7 gives 175 and n=8 gives 232, then 200 lies between two neighbouring terms and cannot be a term. If the equation gives a whole positive n, the number is a term and n is its position.
Method: 1. Substitute the position into the nth term and keep the order of operations. 2. To test a number, compare it with the terms either side of it. 3. Or form the equation, rearrange to zero and solve for n. 4. Accept only a positive whole number for n and say so.
Pattern questions show a sequence of diagrams with dots or tiles. Count the items in the first few patterns, write them as a sequence, and find the nth term in the usual way. The expression then gives the number in any pattern, and you can set it equal to a given number to find which pattern has that many.
Another exam style gives two nth terms and asks for the number that is in both sequences, perhaps between 20 and 40. List the terms of one sequence in the allowed range, then check which of those fit the other. A table of the first few terms of both sequences is quicker than forming an equation, and it shows the working that earns the mark.
Method: 1. Write the first few terms of each sequence in a table. 2. Find the terms of the first sequence that lie in the allowed range. 3. Check each one against the second sequence by solving for n. 4. State the common number and the answer to the question asked.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.