In a geometric sequence each term is the one before it multiplied by the same number, so divide any term by the term before it:

\(16 \div 4 = 4\)

Check it holds all the way along: \(64 \div 16 = 4\) and \(256 \div 64 = 4\).

Keep multiplying by 4:

\(256 \times 4 = 1,024\)

\(1,024 \times 4 = 4,096\)

This is what makes a geometric sequence different from an arithmetic one: the terms are multiplied by 4, not increased by a fixed amount, so the gaps between them grow.

Sequence \(A\) adds 372 each time, so its \(n\)th term is \(372n - 367\).

Sequence \(G\) multiplies by 2 each time, so its \(n\)th term is \(2 \times 2^{n-1}\).

Carry on with both sequences until \(G\) is more than 3716 ahead of \(A\).

At \(n = 12\): \(A = 4097\) and \(G = 4096\), and \(4096 - 4097 = -1\), which is not more than 3716.

At \(n = 13\): \(A = 4469\) and \(G = 8192\), and \(8192 - 4469 = 3723\), which is more than 3716.

Before \(n = 13\) the terms of \(G\) are never above those of \(A\), so this is the first term where the difference is more than 3716.

\(n = 13\)

The first term is 9 and the common ratio is \(27 \div 9 = 3\).

To reach the 1st term you multiply by 3 no times, to reach the 2nd term once, to reach the 3rd term twice — so to reach the \(n\)th term you multiply by 3 exactly \(n - 1\) times.

\(n\)th term \(= 9 \times 3^{n-1}\)

The power is \(n - 1\) and not \(n\); putting \(n = 1\) into the rule has to give back the first term, and \(9 \times 3^{0} = 9\) does.

Put \(n = 7\) into the rule:

\(9 \times 3^{ 6 } = 9 \times 729 = 6,561\)

Divide a term by the term before it:

\(224 \div 896 = \frac{1}{4}\)

Every term is a quarter of the one before it, so the sequence is shrinking. A common ratio between 0 and 1 always gives a decreasing sequence of positive terms.

Multiply the fourth term by the common ratio:

\(14 \times \frac{1}{4} = 14 \div 4 = \frac{7}{2}\)

The terms do not have to stay whole numbers. Dividing 14 by 4 does not go exactly, so the fifth term is the fraction \(\frac{7}{2}\) — and the sequence carries on for ever, getting smaller but never reaching zero.

Each week multiplies the number by 2, so after 2 weeks the starting number has been multiplied by 2 exactly 2 times:

\(32 \times 2^{ 2 } = 128\)

The numbers of ants form a geometric sequence with first term 32 and common ratio 2.

Keep multiplying by 2 and watch for the first number over 8,200.

After 8 weeks: \(32 \times 2^{ 8 } = 8,192\), which is not more than 8,200.

After 9 weeks: \(32 \times 2^{ 9 } = 16,384\), which is.

So the number of ants first passes 8,200 after 9 weeks.

There is no need to work out every term from the start — go straight to a power of 2 that looks about right and then step one either side of it.

The 2nd and 4th terms are two steps apart, so the ratio has been applied twice:

\(r^2 = 4 \div 16 = \frac{1}{4}\)

\(r = \pm \frac{1}{2}\)

All the terms are positive, so the common ratio cannot be negative: \(r = \frac{1}{2}\).

Going back one step divides by \(r\):

first term \(= 16 \div \frac{1}{2} = 32\)

Going forward one step from the 2nd term multiplies by the common ratio \(r = \frac{1}{2}\):

third term \(= 16 \times \frac{1}{2} = 8\)

The common difference is \(4p - p = 3p\), so the third term is

\(4p + 3p = 7p\)

The sum of the first three terms is

\(p + 4p + 7p = 12p\)

This is three times the middle term, \(3 \times 4p\).

\(12p = 144\)

\(p = 144 \div 12 = 12\)

Divide a term by the term before it:

\(-24 \div 8 = -3\)

The signs alternate, which is the giveaway: only a negative common ratio can flip a sequence from positive to negative and back again.

The seventh term is the first term multiplied by the common ratio six times:

\(8 \times (-3)^{6} = 8 \times 729 = 5,832\)

A negative number raised to an even power is positive, so the seventh term is positive — as it must be, because the 1st, 3rd, 5th and 7th terms are all positive.

The 94th term is \(8 \times (-3)^{ 94 - 1 }\), and the sign of that comes entirely from the power of a negative number.

Going along the sequence the signs run positive, negative, positive, negative, … so the odd-numbered terms are positive and the even-numbered terms are negative.

94 is even, so the 94th term is negative.

In a geometric progression each term is the previous term multiplied by the common ratio, so divide the second term by the first.

\(r = \frac{ 9 }{ 3\sqrt{ 3 } }\)

Cancel the common factor and multiply top and bottom by \(\sqrt{ 3 }\) to remove the surd from the denominator:

\(r = \frac{ 3 }{ \sqrt{ 3 } } = \frac{ 3\sqrt{ 3 } }{ 3 } = \sqrt{ 3 }\)

Multiply the third term by the ratio, using \(\sqrt{ 3 } \times \sqrt{ 3 } = 3\):

\(9\sqrt{ 3 } \times \sqrt{ 3 } = 9 \times 3 = 27\)

The third term is \(n = 3\), so cube the bracket. Square it first, using \(\sqrt{ 7 } \times \sqrt{ 7 } = 7\):

\(\left(2 + \sqrt{ 7 }\right)^2 = 4 + 4\sqrt{ 7 } + 7 = 11 + 4\sqrt{ 7 }\)

Multiply by \(2 + \sqrt{ 7 }\), one product at a time:

\(2 \times 11 = 22\) and \(2 \times 4\sqrt{ 7 } = 8\sqrt{ 7 }\)

\(\sqrt{ 7 } \times 11 = 11\sqrt{ 7 }\) and \(\sqrt{ 7 } \times 4\sqrt{ 7 } = 28\)

Collect the whole numbers and the surds separately:

\(22 + 28 = 50\) and \(8\sqrt{ 7 } + 11\sqrt{ 7 } = 19\sqrt{ 7 }\)

\(\left(2 + \sqrt{ 7 }\right)^3 = 50 + 19\sqrt{ 7 }\), the printed third term.

In a geometric sequence each term is the one before it multiplied by the same number, so divide any term by the term before it:

\(40 \div 10 = 4\)

Check it holds all the way along: \(160 \div 40 = 4\) and \(640 \div 160 = 4\).

Keep multiplying by 4:

\(640 \times 4 = 2,560\)

\(2,560 \times 4 = 10,240\)

This is what makes a geometric sequence different from an arithmetic one: the terms are multiplied by 4, not increased by a fixed amount, so the gaps between them grow.

Sequence \(A\) adds 341 each time, so its \(n\)th term is \(341n - 328\).

Sequence \(G\) multiplies by 2 each time, so its \(n\)th term is \(1 \times 2^{n-1}\).

Carry on with both sequences until \(G\) is greater than \(A\).

At \(n = 13\): \(A = 4105\) and \(G = 1 \times 2^{ 12 } = 4096\), so \(G\) is not yet greater.

At \(n = 14\): \(A = 4446\) and \(G = 8192\), so \(G\) is greater.

\(G\) begins below \(A\) and is still not greater at \(n = 13\), so \(n = 14\) is the first term where it is.

\(n = 14\)