NCERT Solutions · Class 9 Mathematics Predicting What Comes Next: Exploring Sequences and Progressions

Substituting into the rule.A sequence given by a rule for \(\displaystyle t_{n}\) is like a machine: feed in a position number \(\displaystyle n\), and out comes the term that sits in that position. "The first five terms" therefore means put \(\displaystyle n = 1, 2, 3, 4, 5\) into the rule, one at a time.(i) \(\displaystyle t_{n} = 3n - 4\)\[t_{1} = 3(1) - 4 = -1, \quad t_{2} = 3(2) - 4 = 2, \quad t_{3} = 3(3) - 4 = 5,\] \[t_{4} = 3(4) - 4 = 8, \quad t_{5} = 3(5) - 4 = 11.\](ii) \(\displaystyle t_{n} = 2 - 5n\)\[t_{1} = 2 - 5(1) = -3, \quad t_{2} = 2 - 5(2) = -8, \quad t_{3} = 2 - 5(3) = -13,\] \[t_{4} = 2 - 5(4) = -18, \quad t_{5} = 2 - 5(5) = -23.\](iii) \(\displaystyle t_{n} = n^{2} - 2n + 3\)\[t_{1} = 1 - 2 + 3 = 2, \quad t_{2} = 4 - 4 + 3 = 3, \quad t_{3} = 9 - 6 + 3 = 6,\] \[t_{4} = 16 - 8 + 3 = 11, \quad t_{5} = 25 - 10 + 3 = 18.\]Putting the three answers side by side:

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$
\(\displaystyle 3n - 4\)	\(\displaystyle -1\)	\(\displaystyle 2\)	\(\displaystyle 5\)	\(\displaystyle 8\)	\(\displaystyle 11\)
\(\displaystyle 2 - 5n\)	\(\displaystyle -3\)	\(\displaystyle -8\)	\(\displaystyle -13\)	\(\displaystyle -18\)	\(\displaystyle -23\)
\(\displaystyle n^{2} - 2n + 3\)	\(\displaystyle 2\)	\(\displaystyle 3\)	\(\displaystyle 6\)	\(\displaystyle 11\)	\(\displaystyle 18\)

Worth noticing. In (i) each term is \(\displaystyle 3\) more than the one before, and in (ii) each is \(\displaystyle 5\) less than the one before — both are arithmetic progressions, and the common difference is exactly the number multiplying \(\displaystyle n\) in the rule. In (iii) the gaps are \(\displaystyle 1, 3, 5, 7\), which keep changing, so that sequence is not an AP.Answer. (i) \(\displaystyle -1,\ 2,\ 5,\ 8,\ 11\) (ii) \(\displaystyle -3,\ -8,\ -13,\ -18,\ -23\) (iii) \(\displaystyle 2,\ 3,\ 6,\ 11,\ 18\)

Substituting into the rule.The rule \(\displaystyle t_{n} = 5n - 3\) works for any position \(\displaystyle n\), so there is no need to write out the terms one by one until we reach the tenth. We simply put \(\displaystyle n = 10\) and \(\displaystyle n = 15\) directly into the rule.For the \(\displaystyle 10^{\text{th}}\) term, take \(\displaystyle n = 10\): \[t_{10} = 5(10) - 3 = 50 - 3 = 47.\]For the \(\displaystyle 15^{\text{th}}\) term, take \(\displaystyle n = 15\): \[t_{15} = 5(15) - 3 = 75 - 3 = 72.\]Check. The sequence starts \(\displaystyle 2,\ 7,\ 12,\ 17,\ \ldots\) — it goes up by \(\displaystyle 5\) each step. From \(\displaystyle t_{1} = 2\), reaching \(\displaystyle t_{10}\) takes \(\displaystyle 9\) steps of \(\displaystyle 5\), so \(\displaystyle t_{10} = 2 + 9 \times 5 = 47\). Going on from \(\displaystyle t_{10}\) to \(\displaystyle t_{15}\) takes \(\displaystyle 5\) more steps: \(\displaystyle 47 + 5 \times 5 = 72\). Both values agree.Answer. \(\displaystyle t_{10} = 47\) and \(\displaystyle t_{15} = 72\).

Working backwards from the rule.A number is a term of this sequence only if some positive whole number \(\displaystyle n\) makes \(\displaystyle 5n - 3\) equal to it. So for each number we set up the equation and see whether the \(\displaystyle n\) that comes out is a counting number. A fractional \(\displaystyle n\) would mean the number falls between two terms and so is not in the sequence at all.Is $\displaystyle 97$ a term? Put \(\displaystyle t_{n} = 97\): \[5n - 3 = 97 \;\Rightarrow\; 5n = 100 \;\Rightarrow\; n = 20.\] Here \(\displaystyle n = 20\) is a positive whole number, so yes. Checking: \(\displaystyle 5(20) - 3 = 100 - 3 = 97\). ✓Is $\displaystyle 172$ a term? Put \(\displaystyle t_{n} = 172\): \[5n - 3 = 172 \;\Rightarrow\; 5n = 175 \;\Rightarrow\; n = 35.\] Again \(\displaystyle n = 35\) is a positive whole number, so yes. Checking: \(\displaystyle 5(35) - 3 = 175 - 3 = 172\). ✓Another way to see it. Every term \(\displaystyle 5n - 3\) leaves remainder \(\displaystyle 2\) on division by \(\displaystyle 5\) (since \(\displaystyle 5n - 3 = 5(n-1) + 2\)). Both \(\displaystyle 97 = 5 \times 19 + 2\) and \(\displaystyle 172 = 5 \times 34 + 2\) leave remainder \(\displaystyle 2\), and both are bigger than the first term \(\displaystyle 2\), so both must appear. A number such as \(\displaystyle 100\) or \(\displaystyle 96\), which leaves a different remainder, could never be a term.Answer. Both are terms: \(\displaystyle 97\) is the \(\displaystyle 20^{\text{th}}\) term and \(\displaystyle 172\) is the \(\displaystyle 35^{\text{th}}\) term.

Working backwards from the rule.We are told the value of a term and asked for its position. So put \(\displaystyle t_{n} = 607\) and solve for \(\displaystyle n\): \[5n - 3 = 607.\]Add \(\displaystyle 3\) to both sides: \[5n = 610.\]Divide both sides by \(\displaystyle 5\): \[n = 122.\]Since \(\displaystyle 122\) is a positive whole number, \(\displaystyle 607\) really does occupy a position in this sequence.Check. \(\displaystyle t_{122} = 5(122) - 3 = 610 - 3 = 607\). ✓Answer. \(\displaystyle 607\) is the \(\displaystyle 122^{\text{nd}}\) term.

Recursive rule, then a closed form.A recursive rule tells you where to start and how to take one step. Here the start is \(\displaystyle t_{1} = -5\), and the step \(\displaystyle t_{n+1} = t_{n} + 3\) says: to get the next term, add \(\displaystyle 3\) to the one you have.The first five terms. Adding \(\displaystyle 3\) each time: \[t_{1} = -5, \quad t_{2} = -5 + 3 = -2, \quad t_{3} = -2 + 3 = 1, \quad t_{4} = 1 + 3 = 4, \quad t_{5} = 4 + 3 = 7.\]So the sequence begins \(\displaystyle -5,\ -2,\ 1,\ 4,\ 7,\ \ldots\) — an AP with first term \(\displaystyle -5\) and common difference \(\displaystyle 3\).Is $\displaystyle 52$ a term? Stepping up by \(\displaystyle 3\) all the way to \(\displaystyle 52\) would take a long time, so first turn the recursive rule into a direct one. Going from \(\displaystyle t_{1}\) to \(\displaystyle t_{n}\) takes \(\displaystyle (n-1)\) steps, and each step adds \(\displaystyle 3\): \[t_{n} = -5 + (n-1)\times 3 = 3n - 8.\]Now set this equal to \(\displaystyle 52\): \[3n - 8 = 52 \;\Rightarrow\; 3n = 60 \;\Rightarrow\; n = 20.\]\(\displaystyle n = 20\) is a positive whole number, so \(\displaystyle 52\) does occur.Check. \(\displaystyle t_{20} = 3(20) - 8 = 60 - 8 = 52\). ✓ (Also, \(\displaystyle t_{5} = 3(5) - 8 = 7\), which matches the fifth term found above, so the closed form is the right one.)Answer. The first five terms are \(\displaystyle -5,\ -2,\ 1,\ 4,\ 7\). Yes, \(\displaystyle 52\) is a term — it is the \(\displaystyle 20^{\text{th}}\) term.

Building a recursive sequence step by step.This rule needs three earlier terms to make a new one: each term from the fourth onwards is the sum of the three terms immediately before it. That is why the book has to hand us three starting values, \(\displaystyle \mathrm{T}_{1} = 1\), \(\displaystyle \mathrm{T}_{2} = 2\), \(\displaystyle \mathrm{T}_{3} = 4\). We just apply the rule \(\displaystyle \mathrm{T}_{n} = \mathrm{T}_{n-1} + \mathrm{T}_{n-2} + \mathrm{T}_{n-3}\) one term at a time, always using the three most recent values.\[\mathrm{T}_{4} = \mathrm{T}_{3} + \mathrm{T}_{2} + \mathrm{T}_{1} = 4 + 2 + 1 = 7,\] \[\mathrm{T}_{5} = \mathrm{T}_{4} + \mathrm{T}_{3} + \mathrm{T}_{2} = 7 + 4 + 2 = 13,\] \[\mathrm{T}_{6} = \mathrm{T}_{5} + \mathrm{T}_{4} + \mathrm{T}_{3} = 13 + 7 + 4 = 24,\] \[\mathrm{T}_{7} = \mathrm{T}_{6} + \mathrm{T}_{5} + \mathrm{T}_{4} = 24 + 13 + 7 = 44,\] \[\mathrm{T}_{8} = \mathrm{T}_{7} + \mathrm{T}_{6} + \mathrm{T}_{5} = 44 + 24 + 13 = 81.\]Laid out in order:

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle \mathrm{T}_{n}\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 4$	$\displaystyle 7$	$\displaystyle 13$	$\displaystyle 24$	$\displaystyle 44$	$\displaystyle 81$

Notice that the sequence is neither an AP (the differences \(\displaystyle 1, 2, 3, 6, 11, 20, 37\) are not constant) nor a GP (the ratios are not constant either). It is a close cousin of the Fibonacci sequence, which adds the previous two terms instead of three.Answer. \(\displaystyle \mathrm{T}_{4} = 7,\ \mathrm{T}_{5} = 13,\ \mathrm{T}_{6} = 24,\ \mathrm{T}_{7} = 44,\ \mathrm{T}_{8} = 81\).