The gaps between the terms are not equal, so the sequence does not go up by a fixed amount. Look at how the gaps themselves change.

First differences: \(17,\ 27,\ 37\)

Second differences: \(10,\ 10\)

The second difference stays at \(10\), so the next gap is \(10\) more than the last one:

next gap \(= 37 + 10 = 47\), so the fifth term is \(82 + 47 = 129\)

the gap after that \(= 47 + 10 = 57\), so the sixth term is \(129 + 57 = 186\)

The second difference is the amount the gaps go up by, and it is the same all the way along: \(10\).

It is worth remembering what it is for. The second difference is always twice the coefficient of \(n^2\), so this sequence has \(5n^2\) in its rule.

Put \(n = 1\), then \(n = 2\), then \(n = 3\) into the expression, squaring the \(n\) before multiplying.

\(n = 1:\quad 3 \times 1^2 + 7 \times 1 - 9 = 1\)

\(n = 2:\quad 3 \times 2^2 + 7 \times 2 - 9 = 17\)

\(n = 3:\quad 3 \times 3^2 + 7 \times 3 - 9 = 39\)

\(1,\ 17,\ 39\)

Worth a check before moving on: the second difference of a quadratic sequence is always double the number in front of \(n^2\). Here that is \(6\), and the two gaps between these three terms differ by exactly that much.

In a quadratic sequence the second difference is the same all the way along. Call it \(D\).

Two of the gaps can be worked out straight away:

first gap \(= 37 - 19 = 18\)

last gap \(= 151 - 103 = 48\)

There are four gaps altogether, so getting from the first to the last one means adding \(D\) three times:

\(18 + 3D = 48\)

\(3D = 30\), so \(D = 10\)

Now walk forwards. The second gap is \(18 + 10 = 28\), so the missing term is

\(37 + 28 = 65\)

Check it against the other end: the third gap should be \(28 + 10 = 38\), and \(65 + 38 = 103\), which is the fourth term the question gave. It fits.

Work out the differences between consecutive terms of each sequence, and then the differences between those.

Sequence \(A\) has first differences \(15,\ 23,\ 31,\ 39\), which are not all the same, so it is not linear. Its second differences are \(8,\ 8,\ 8\) — all equal and not zero, and that is exactly what makes a sequence quadratic.

The other two fail the test in the two ways a sequence can. One goes up by the same amount every time, so its first differences are constant and its second differences are all zero — that is a linear sequence. The other multiplies by the same number every time, so its differences keep growing and never settle.

Work out the gaps between the terms, and then the gaps between the gaps.

First differences: \(11,\ 17,\ 23,\ 29\)

Second differences: \(6,\ 6,\ 6\)

They are all the same, which is what makes this a quadratic sequence. The second difference is always double the number in front of \(n^2\), so

\(\text{coefficient of } n^2 = 6 \div 2 = 3\)

Now take \(3n^2\) away from every term. Line up the two rows:

\(3n^2\) gives \(3,\ 12,\ 27,\ 48,\ 75\)

the sequence is \(5,\ 16,\ 33,\ 56,\ 85\)

so what is left over is \(2,\ 4,\ 6,\ 8,\ 10\)

That leftover sequence changes by the same amount every time, so it is an ordinary linear one. It steps up by \(2\) each time, and working back one step from \(2\) gives \(0\) at position zero, so its \(n\)th term is \(2n \).

Add the two pieces back together:

\(n\text{th term} = 3n^2 + 2n \)

First differences: \(12,\ 16,\ 20,\ 24\)

Second differences: \(4,\ 4,\ 4\)

The second difference is constant, so the sequence is quadratic, and halving it gives the coefficient of \(n^2\):

\(4 \div 2 = 2\)

Subtract \(2n^2\) from each term. The squares scaled by \(2\) are \(2,\ 8,\ 18,\ 32,\ 50\), and taking them off \(7,\ 19,\ 35,\ 55,\ 79\) leaves

\(5,\ 11,\ 17,\ 23,\ 29\)

which goes up by \(6\) each time and so has \(n\)th term \(6n - 1\).

\(n\text{th term} = 2n^2 + 6n - 1\)

Working along the sequence as far as the 12th term would be slow, and one slip on the way would spoil everything after it. Put \(n = 12\) into the rule instead.

\(12^2 = 144\), so the \(n^2\) part is \(2 \times 144 = 288\)

The \(n\) part is \(6 \times 12 = 72\)

\(288 + 72 - 1 = 359\)

Squaring comes before multiplying, so it is \(2 \times 12^2\) and never \((2 \times 12)^2\).

"Which term is \(83\)" means "what value of \(n\) makes the \(n\)th term equal to \(83\)", so set the expression equal to it and solve.

\(n^2 + 3n - 5 = 83\)

Take \(83\) across so that one side is zero:

\(n^2 + 3n - 88 = 0\)

Factorise:

\((n - 8)(n + 11) = 0\)

\(n = 8\) or \(n = -11\)

A position has to be a positive whole number, so \(n = -11\) is not one — the sequence has no \(-11\)th term. That leaves

\(n = 8\)

Check it: \(8^2 + 3 \times 8 - 5 = 83\).

The second difference of a quadratic sequence is always double the number in front of \(n^2\), so \(a\) is half of it.

\(a = 2 \div 2 = 1\)

Put \(n = 1\) and \(n = 2\) into \(n^2 + bn + c\) and set each equal to the term the question gives.

\(n = 1:\quad 1 + b + c = 18\)

\(n = 2:\quad 4 + 2b + c = 23\)

Subtract the first equation from the second. The \(c\) disappears:

\(3 + b = 5\)

\(b = 2\)

Now put that back into the first equation:

\(1 + 2 + c = 18\)

\(c = 15\)

So the \(n\)th term is \(n^2 + 2n + 15\).

Every term of this sequence is bigger than the one before it, so once the terms pass \(33\) they stay past it. That means there is a first one, and the job is to find where the sequence crosses.

Try \(n = 3\):

\(9 + 12 + 9 = 30\)

That is not greater than \(33\).

Try \(n = 4\):

\(16 + 16 + 9 = 41\)

That is greater than \(33\), and it is the first one that is, because the term before it was not.

The first term greater than \(33\) is \(41\).

Note what the question asks for: the term itself, \(41\), not its position \(4\).

Find the rule first. The first differences are \(7,\ 11,\ 15\) and the second differences are \(4,\ 4\), so the sequence is quadratic with \(2n^2\) in it, and after taking \(2n^2\) off each term what is left goes up by \(1\) each time.

\(n\text{th term} = 2n^2 +n - 3\)

Now put values of \(n\) into the rule and work along the sequence until you reach 133 or pass it. The two terms that straddle it are these:

\(n = 8\): \(2 \times 64 + 8 - 3 = 133\)

\(n = 9\): \(2 \times 81 + 9 - 3 = 168\)

Compare 133 with those two terms. It is exactly equal to one of them.

Yes — 133 is a term of this sequence.