Predicting What Comes Next: Exploring Sequences and Progressions

An explicit formula tells you a term straight from its position number. So put \(n = 1, 2, 3, 4, 5\) into each rule in turn.

(i) \(t_n = 3n - 4\).

First five terms: −1, 2, 5, 8, 11. The difference row is a constant 3, so this is an AP with \(d = 3\) — which you could have read straight off the \(3n\).

(ii) \(t_n = 2 - 5n\).

First five terms: −3, −8, −13, −18, −23. Constant difference −5, so this is a decreasing AP.

(iii) \(t_n = n^2 - 2n + 3\).

First five terms: 2, 3, 6, 11, 18.

The formula is explicit, so no earlier terms are needed — just substitute.

\(t_{10} = \mathbf{47}\) and \(t_{15} = \mathbf{72}\).

The first few terms above show the pattern: this is an AP starting at 2 with a common difference of 5.

A number is a term of this sequence exactly when the equation \(5n - 3 = \text{that number}\) has a solution \(n\) that is a positive whole number. So solve, then check what kind of number you got.

Is 97 a term?

20 is a positive whole number, so yes — 97 is the \(20^{th}\) term.

Is 172 a term?

35 is also a positive whole number, so yes — 172 is the \(35^{th}\) term.

Set the formula equal to 607 and solve for the position \(n\).

So 607 is the \(122^{nd}\) term.

Check: \(t_{122} = 5(122) - 3 = 610 - 3 = 607\). ✓

A recursive rule builds each term out of the one before it. Start from \(t_1 = -5\) and add 3 over and over.

First five terms: −5, −2, 1, 4, 7.

The difference row is a constant 3, so this is an AP with \(a = -5\) and \(d = 3\). That lets us write an explicit formula, which is far quicker than stepping term by term.

Is 52 a term?

Yes — 52 is the \(20^{th}\) term.

This recursion reaches back three terms instead of one, so each new term is the sum of the previous three. Work downwards, keeping the three most recent values in front of you.

So \(T_4 = \mathbf{7}\), \(T_5 = \mathbf{13}\), \(T_6 = \mathbf{24}\), \(T_7 = \mathbf{44}\), \(T_8 = \mathbf{81}\).

Reading the sequence: first term \(a = 3\). Subtract each term from the next to find the common difference.

The difference row is a constant 5, which confirms it really is an AP, with \(a = 3\) and \(d = 5\).

The \(10^{th}\) term:

The \(26^{th}\) term:

So \(t_{10} = \mathbf{48}\) and \(t_{26} = \mathbf{128}\).

Reading the sequence: \(a = 21\), and each term is 3 less than the one before, so \(d = -3\).

Which term is −81?

−81 is the \(35^{th}\) term.

Is 0 a term? Ask the same question of 0: is there a positive whole number \(n\) that makes \(t_n = 0\)?

\(n = 8\) is a positive whole number, so yes — 0 is the \(8^{th}\) term.

Reason, written out: the terms run 21, 18, 15, 12, 9, 6, 3, 0, −3, ... . Starting from 21 and stepping down by 3 keeps every term a multiple of 3, and since 21 is itself a multiple of 3, the sequence lands exactly on 0 rather than stepping over it.

Reading the sequence: \(a = 11\); each term is 3 less than the last, so \(d = -3\).

The explicit formula:

Check: \(t_1 = 14 - 3 = 11\) ✓ and \(t_4 = 14 - 12 = 2\) ✓.

The recursive rule. A recursion needs two things: where to start, and how to get from one term to the next.

So the \(n^{th}\) term is \(\mathbf{t_n = 14 - 3n}\), and the recursive rule is \(t_1 = 11,\ t_{n+1} = t_n - 3\).

50 terms means the last one is \(t_{50}\). Write both pieces of information as equations in \(a\) and \(d\), exactly as the hint suggests.

Subtract the first equation from the second — the \(a\) cancels.

Substitute back into \(a + 2d = 12\):

Check both given facts: \(t_3 = 8 + 4 = 12\) ✓ and \(t_{50} = 8 + 98 = 106\) ✓.

Now the \(29^{th}\) term:

The \(29^{th}\) term is 64.

Set it up as an AP. The smallest two-digit multiple of 3 is 12 (since 9 is one digit and 10, 11 are not multiples of 3); the largest is 99. Between them the multiples step up by 3.

So \(a = 12\), \(d = 3\) and the last term is 99.

How many terms?

There are 30 two-digit multiples of 3.

Their sum. Use Āryabhaṭa’s form of the rule: the sum is the average of the first and last terms, multiplied by the number of terms.

The sum of all two-digit multiples of 3 is 1665.

The yearly salaries form an AP: the first year’s salary is \(a = 500000\) and each later year adds \(d = 20000\).

Find which year’s salary is ₹7,00,000.

So ₹7,00,000 is the salary in the \(11^{th}\) year of work, which is reached after 10 years of increments.

Row \(n\) holds \(n\) marbles, so the total is the sum of the first 25 natural numbers.

This is an AP with \(a = 1\), \(d = 1\) and \(n = 25\) terms, so use the sum rule.

The child uses 325 marbles in all.

There are two clean routes. The short one uses the fact that in a GP, moving forward 4 places multiplies by \(r^4\).

Short route. From the \(8^{th}\) term to the \(12^{th}\) is 4 steps:

Long route, via the first term.

Both routes give \(t_{12} = \mathbf{3072}\).

The ratio row is a constant 2, confirming the first term \(a = \tfrac{3}{2}\) is consistent with the given common ratio.

Reading the sequence: \(a = 5\), and each term is 5 times the one before it, so \(r = 5\).

The \(n^{th}\) term:

The \(10^{th}\) term:

So \(t_n = \mathbf{5^n}\) and \(t_{10} = \mathbf{9\,765\,625}\).

Generate the terms first, and watch what the difference and ratio rows do.

The terms themselves are neither an AP nor a GP — but look at the difference row: 2, 6, 18, 54 is a GP with ratio 3. That is the clue.

Subtract 1 from every term and the pattern falls out:

Those are the powers of 3, so:

Check: \(t_1 = 3^0 + 1 = 2\) ✓, \(t_5 = 81 + 1 = 82\) ✓.

Now find 730:

730 is the \(7^{th}\) term.

Reading the sequence: \(a = 2\), and \(6 \div 2 = 3\), \(18 \div 6 = 3\), so \(r = 3\).

Explicit formula:

Recursive formula:

Which term is 4374?

4374 is the \(8^{th}\) term.

Check: \(2 \times 3^7 = 2 \times 2187 = 4374\) ✓.

The peak heights after each bounce form a GP with first term \(80 \times 0.6 = 48\) and common ratio \(r = 0.6\).

(i) The height after the \(n^{th}\) bounce is \(80 \times (0.6)^n\).

The ball reaches 6.2208 m after the \(5^{th}\) bounce — about 6.22 m.

(ii) Count the journey carefully. The ball falls 80 m to hit the ground the 1st time. After that, each rise is followed by an equal fall, and each of those falls ends in another ground contact.

So the total is the first 80 m plus twice the sum of the first five rebound heights.

The total vertical distance is 301.3376 m, about 301.34 m.

Reading the sequence: divide each term by the one before it.

The ratio is the same both times, so this is a GP with \(a = 2\) and \(r = \sqrt{2}\).

Everything here is a power of 2, so rewrite it that way — it makes the equation trivial.

Now set that equal to \(128 = 2^7\):

128 is the \(13^{th}\) term.

Check: \(t_{13} = 2 \times (\sqrt{2})^{12} = 2 \times 2^6 = 2 \times 64 = 128\) ✓.

Reading the figure: at each stage every red square is cut into 9 equal squares and the middle one is thrown away, leaving 8. So one red square becomes eight red squares, each with \(\tfrac{1}{9}\) of its area.

(i) Count stage by stage: Stage 0 is a single square; each later stage has 8 times as many.

(ii) The ratio row is a constant 8, so just keep multiplying.

(iii) The counts form a GP with first term 1 (at stage 0) and common ratio 8.

Explicit formula:

Recursive formula:

(iv) Now the area. Stage 0 has area 1. Each stage keeps 8 of the 9 equal pieces, so the area is multiplied by \(\tfrac{8}{9}\) every time — another GP, but a shrinking one.

Explicit formula:

Recursive formula:

What happens as \(n\) increases? Since \(\tfrac{8}{9}\) is less than 1, every stage makes the area smaller, and it keeps shrinking without ever stopping. The area of the red region tends to 0 as \(n\) grows.

Write both facts as equations in \(a\) and \(d\).

Subtract to eliminate \(a\):

Substitute back:

Check both given facts: \(t_{11} = -32 + 70 = 38\) ✓ and \(t_{16} = -32 + 105 = 73\) ✓.

The \(31^{st}\) term is 178.

"Determine the AP" means find \(a\) and \(d\), then write the sequence out.

For the second fact, subtract the \(5^{th}\) term from the \(7^{th}\):

Substitute into the first equation:

The AP is 4, 10, 16, 22, 28, ...

Check: the third term is 16 ✓, and \(t_7 = 40\), \(t_5 = 28\), so \(t_7 - t_5 = 12\) ✓.

Find the first and last such numbers. \(7 \times 14 = 98\), which has two digits, and \(7 \times 15 = 105\), so 105 is the smallest three-digit multiple of 7. At the top, \(7 \times 142 = 994\) and \(7 \times 143 = 1001\), so 994 is the largest.

These form an AP with \(a = 105\) and \(d = 7\). Find how many terms it has.

There are 128 three-digit numbers divisible by 7.

Find the first and last such numbers. The first multiple of 4 above 10 is 12; the last one below 250 is 248 (since \(4 \times 62 = 248\) and \(4 \times 63 = 252\)).

An AP with \(a = 12\) and \(d = 4\).

There are 60 multiples of 4 between 10 and 250.

Check: the multiples are \(4 \times 3\) up to \(4 \times 62\), and \(62 - 3 + 1 = 60\) ✓.

Let the GP be \(a, ar, ar^2, \dots\) and turn each sentence into an equation.

Start with the second equation. Since \(a \ne 0\) and \(r \ne 0\) (or there would be no GP at all), divide both sides by \(ar^2\):

Both signs are legitimate, so there are two GPs. Take each in turn.

Case 1: \(r = 2\).

Case 2: \(r = -2\).

Check Case 2: first two terms \(4 + (-8) = -4\) ✓; third term 16, fifth term 64, and \(64 = 4 \times 16\) ✓.

Suppose the sum runs over \(k\) consecutive numbers starting at \(a\). It is an AP with \(d = 1\), so:

So \(k\) must be a factor of 200, and for each factor we test whether \(a\) comes out as a natural number. One useful observation cuts the work down: \(k\) and \(2a + k - 1\) always have opposite parity (one odd, one even), so one of the two factors must be an odd factor of 200. The odd factors of \(200 = 2^3 \times 5^2\) are 1, 5 and 25.

So there are exactly two ways.

Check the first: five terms averaging 20, so \(5 \times 20 = 100\) ✓. Check the second: eight terms averaging 12.5, so \(8 \times 12.5 = 100\) ✓.

Doubling every hour is a GP with common ratio 2, starting from 30 bacteria at hour 0.

End of the \(2^{nd}\) hour:

End of the \(4^{th}\) hour:

End of the \(n^{th}\) hour:

So there are 120 bacteria after 2 hours, 480 after 4 hours, and \(\mathbf{30 \times 2^n}\) after \(n\) hours.

Write each sum in terms of \(a\) and \(d\).

Subtract the first simplified equation from the second:

The first three terms are −13, −8, −3.

Check both given facts: \(t_4 = 2\) and \(t_8 = 22\), so their sum is 24 ✓; \(t_6 = 12\) and \(t_{10} = 32\), so their sum is 44 ✓.

The sum of the first \(n\) natural numbers is the \(n^{th}\) triangular number.

We want the smallest \(n\) with \(S_n\) greater than 1000, so we need:

Since \(44^2 = 1936\) and \(45^2 = 2025\), the answer is near 44. Test both neighbours:

1980 is not more than 2000 but 2070 is, so \(n = 45\).

The smallest such value is \(n = 45\).

Reading the sequence: \(8 \div 2 = 4\) and \(32 \div 8 = 4\), so \(a = 2\) and \(r = 4\).

Explicit formula:

Recursive formula:

Which term is 131072?

131072 is the \(9^{th}\) term.

Check: \(4^8 = 65536\) and \(2 \times 65536 = 131072\) ✓.

When a question gives you the product of three GP terms, write them as \(\frac{a}{r},\ a,\ ar\) rather than \(a,\ ar,\ ar^2\). The \(r\) then cancels out of the product and the work collapses.

Now use the sum, with \(a = -1\):

Multiply through by \(12r\):

Solve the quadratic:

Take \(r = -\frac{3}{4}\). The three terms \(\frac{a}{r},\ a,\ ar\) are:

So the terms are \(\frac{4}{3},\ -1,\ \frac{3}{4}\) with common ratio \(-\frac{3}{4}\).

Check the sum:

Check the product:

Let the GP have first term \(a\) and common ratio \(r\), so \(t_n = ar^{\,n-1}\). Write down the three given terms.

Three numbers are in GP exactly when the middle one squared equals the product of the outer two — that is, when \(y^2 = xz\). So compute both sides.

The two sides are identical, so \(y^2 = xz\) and therefore \(x, y, z\) are in GP. ■

The common ratio of the new GP:

So \(x, y, z\) form a GP with common ratio \(r^6\).

Let the terms be \(a,\ ar,\ ar^2\).

The second expression factorises in a way that is worth remembering:

So the second equation becomes:

But \(a(1 + r + r^2) = 26\), so substitute that in:

Now divide the two boxed results to knock out \(a\) altogether:

With \(r = 3\):

The terms are 2, 6, 18.

Check: \(2 + 6 + 18 = 26\) ✓ and \(4 + 36 + 324 = 364\) ✓.

Each new term is one more than the sum of everything before it. Build them up one at a time, keeping a running total.

So \(P_1\) to \(P_8\) are 1, 2, 4, 8, 16, 32, 64, 128.

A simpler recursive formula. The ratio row is a constant 2, so each term is simply double the last. Here is why, in one line: \(P_n\) is the sum of everything before it plus 1, and \(P_{n-1}\) is the sum of everything before it plus 1, so subtracting gives

An explicit formula. The sequence is a GP with first term 1 and common ratio 2.

Check: \(P_1 = 2^0 = 1\) ✓, \(P_8 = 2^7 = 128\) ✓.

This looks like the previous question, but read the last subscript carefully: the sum stops at \(W_{n-2}\), not \(W_{n-1}\), and the number added is 2 rather than 1. Those two changes make all the difference.

So \(W_1\) to \(W_8\) are 1, 2, 3, 5, 8, 13, 21, 34.

Do you recognise it? Yes — this is the Virahāṅka–Fibonacci sequence met earlier in this chapter, where \(V_1 = 1\), \(V_2 = 2\) and each term is the sum of the two before it.

Why the two rules agree. Compare \(W_n\) with \(W_{n-1}\):

Subtracting, everything cancels except the one extra term in the longer sum:

which is precisely the Virahāṅka–Fibonacci rule. ■