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

\(\displaystyle n^{\text{th}}\) term of an AP.First check that this really is an AP by looking at consecutive differences: \[8 - 3 = 5, \qquad 13 - 8 = 5, \qquad 18 - 13 = 5.\] The difference is the same every time, so it is an AP with first term \(\displaystyle a = 3\) and common difference \(\displaystyle d = 5\).To reach the \(\displaystyle n^{\text{th}}\) term from the first, you take \(\displaystyle (n-1)\) steps of size \(\displaystyle d\) — not \(\displaystyle n\) steps, because the first term is already there before any step is taken. That gives the standard rule \[t_{n} = a + (n-1)d = 3 + (n-1)\times 5 = 5n - 2.\]The \(\displaystyle 10^{\text{th}}\) term: \[t_{10} = 3 + (10 - 1)\times 5 = 3 + 45 = 48.\]The \(\displaystyle 26^{\text{th}}\) term: \[t_{26} = 3 + (26 - 1)\times 5 = 3 + 125 = 128.\]Check with the short rule. \(\displaystyle 5(10) - 2 = 48\) and \(\displaystyle 5(26) - 2 = 128\). ✓ Also \(\displaystyle 5(1) - 2 = 3\) and \(\displaystyle 5(4) - 2 = 18\), matching the first and fourth listed terms.Answer. \(\displaystyle t_{10} = 48\) and \(\displaystyle t_{26} = 128\).

\(\displaystyle n^{\text{th}}\) term of an AP, then solving for the position.The differences are \(\displaystyle 18 - 21 = -3\) and \(\displaystyle 15 - 18 = -3\), so this is an AP with \(\displaystyle a = 21\) and \(\displaystyle d = -3\). A negative common difference simply means the terms are going down. Its \(\displaystyle n^{\text{th}}\) term is \[t_{n} = a + (n-1)d = 21 + (n-1)(-3) = 21 - 3n + 3 = 24 - 3n.\]Which term is \(\displaystyle -81\)? Set \(\displaystyle t_{n} = -81\): \[24 - 3n = -81 \;\Rightarrow\; -3n = -105 \;\Rightarrow\; n = 35.\] Since \(\displaystyle 35\) is a positive whole number, \(\displaystyle -81\) is genuinely a term. Check: \(\displaystyle t_{35} = 24 - 3(35) = 24 - 105 = -81\). ✓Is \(\displaystyle 0\) a term? Set \(\displaystyle t_{n} = 0\): \[24 - 3n = 0 \;\Rightarrow\; 3n = 24 \;\Rightarrow\; n = 8.\] Again \(\displaystyle n = 8\) is a positive whole number, so yes, \(\displaystyle 0\) is a term. The reason it works is that \(\displaystyle 24\) happens to be exactly divisible by \(\displaystyle 3\); had the equation given a fraction such as \(\displaystyle n = 8.5\), the value \(\displaystyle 0\) would have fallen between two terms and would not be in the AP.We can see it directly by writing the AP out: \[21,\ 18,\ 15,\ 12,\ 9,\ 6,\ 3,\ \mathbf{0},\ -3,\ \ldots\] and \(\displaystyle 0\) is indeed in the eighth place. ✓Answer. \(\displaystyle -81\) is the \(\displaystyle 35^{\text{th}}\) term. Yes, \(\displaystyle 0\) is a term of this AP — it is the \(\displaystyle 8^{\text{th}}\) term, because \(\displaystyle 24 - 3n = 0\) gives the whole number \(\displaystyle n = 8\).

\(\displaystyle n^{\text{th}}\) term of an AP, plus its recursive rule.Check the differences first: \[8 - 11 = -3, \qquad 5 - 8 = -3, \qquad 2 - 5 = -3.\] Constant, so this is an AP with first term \(\displaystyle a = 11\) and common difference \(\displaystyle d = -3\).Explicit (\(\displaystyle n^{\text{th}}\)-term) rule. From the first term you take \(\displaystyle (n-1)\) steps of \(\displaystyle -3\) to arrive at the \(\displaystyle n^{\text{th}}\) term: \[t_{n} = a + (n-1)d = 11 + (n-1)(-3) = 11 - 3n + 3 = 14 - 3n.\]Check it on the given terms: \(\displaystyle t_{1} = 14 - 3 = 11\), \(\displaystyle t_{2} = 14 - 6 = 8\), \(\displaystyle t_{3} = 14 - 9 = 5\), \(\displaystyle t_{4} = 14 - 12 = 2\). ✓ All four match.Recursive rule. A recursive rule says where to start and how to take one step. Here you start at \(\displaystyle 11\) and each new term is \(\displaystyle 3\) less than the previous one: \[t_{1} = 11, \qquad t_{n+1} = t_{n} - 3 \quad \text{for } n \geq 1.\]The two rules describe exactly the same sequence. The recursive one is handy for writing the next few terms; the explicit one is what you want when someone asks for the \(\displaystyle 100^{\text{th}}\) term without listing the other 99.Answer. \(\displaystyle t_{n} = 14 - 3n\); recursive rule: \(\displaystyle t_{1} = 11,\ t_{n+1} = t_{n} - 3\).

Two unknowns, two equations.Every term of an AP is built from just two numbers, the first term \(\displaystyle a\) and the common difference \(\displaystyle d\), through \(\displaystyle t_{n} = a + (n-1)d\). So two pieces of information are enough to pin the whole AP down.The \(\displaystyle 3^{\text{rd}}\) term is \(\displaystyle 12\): \[a + (3-1)d = 12 \;\Rightarrow\; a + 2d = 12. \qquad \ldots (1)\]The AP has \(\displaystyle 50\) terms, so the last term is the \(\displaystyle 50^{\text{th}}\), and it is \(\displaystyle 106\): \[a + (50-1)d = 106 \;\Rightarrow\; a + 49d = 106. \qquad \ldots (2)\]Solve the pair. Subtracting \(\displaystyle (1)\) from \(\displaystyle (2)\) removes \(\displaystyle a\) altogether: \[(a + 49d) - (a + 2d) = 106 - 12 \;\Rightarrow\; 47d = 94 \;\Rightarrow\; d = 2.\]Put \(\displaystyle d = 2\) back into \(\displaystyle (1)\): \[a + 2(2) = 12 \;\Rightarrow\; a + 4 = 12 \;\Rightarrow\; a = 8.\]Now the \(\displaystyle 29^{\text{th}}\) term: \[t_{29} = a + (29-1)d = 8 + 28 \times 2 = 8 + 56 = 64.\]Check. With \(\displaystyle a = 8,\ d = 2\): the \(\displaystyle 3^{\text{rd}}\) term is \(\displaystyle 8 + 4 = 12\) ✓ and the \(\displaystyle 50^{\text{th}}\) term is \(\displaystyle 8 + 49 \times 2 = 8 + 98 = 106\) ✓. Both given facts are reproduced.Answer. \(\displaystyle a = 8\), \(\displaystyle d = 2\), and the \(\displaystyle 29^{\text{th}}\) term is \(\displaystyle 64\).

Counting the terms of an AP, then adding them.The \(\displaystyle 2\)-digit numbers run from \(\displaystyle 10\) to \(\displaystyle 99\). The smallest of these divisible by \(\displaystyle 3\) is \(\displaystyle 12\), and the largest is \(\displaystyle 99\) (since \(\displaystyle 9 + 9 = 18\) is divisible by \(\displaystyle 3\)). Between them the multiples of \(\displaystyle 3\) go up in steps of \(\displaystyle 3\), so they form the AP \[12,\ 15,\ 18,\ \ldots,\ 99, \qquad a = 12,\quad d = 3.\]How many are there? Let \(\displaystyle 99\) be the \(\displaystyle n^{\text{th}}\) term: \[a + (n-1)d = 99 \;\Rightarrow\; 12 + (n-1)3 = 99 \;\Rightarrow\; (n-1)3 = 87 \;\Rightarrow\; n - 1 = 29 \;\Rightarrow\; n = 30.\] So there are \(\displaystyle 30\) such numbers.Their sum. Pair the first with the last, the second with the second-last, and so on: \(\displaystyle 12 + 99 = 111\), \(\displaystyle 15 + 96 = 111\), \(\displaystyle 18 + 93 = 111\), … every pair adds to the same total. With \(\displaystyle 30\) numbers there are \(\displaystyle 15\) such pairs, so \[\text{Sum} = 15 \times 111 = 1665.\]This pairing is exactly the standard formula \(\displaystyle S_{n} = \dfrac{n}{2}(a + l)\), where \(\displaystyle l\) is the last term: \[S_{30} = \frac{30}{2}(12 + 99) = 15 \times 111 = 1665.\]Check a different way. The multiples of \(\displaystyle 3\) from \(\displaystyle 12\) to \(\displaystyle 99\) are \(\displaystyle 3(4 + 5 + \cdots + 33)\). Now \(\displaystyle 4 + 5 + \cdots + 33 = \dfrac{30(4 + 33)}{2} = 15 \times 37 = 555\), and \(\displaystyle 3 \times 555 = 1665\). ✓ The two routes agree.Answer. There are \(\displaystyle 30\) two-digit numbers divisible by \(\displaystyle 3\), and their sum is \(\displaystyle 1665\).

Salary as an AP.The salary rises by the same amount, ₹$\displaystyle 20,000$, every year, so the yearly salaries form an AP with \[a = 5{,}00{,}000 \quad \text{(salary in the 1st year)}, \qquad d = 20{,}000.\]The salary in the \(\displaystyle n^{\text{th}}\) year is \[t_{n} = a + (n-1)d = 5{,}00{,}000 + (n-1)\times 20{,}000.\]Set this equal to ₹$\displaystyle 7,00,000$: \[5{,}00{,}000 + (n-1)\times 20{,}000 = 7{,}00{,}000.\]Subtract \(\displaystyle 5{,}00{,}000\) from both sides: \[(n-1)\times 20{,}000 = 2{,}00{,}000.\]Divide both sides by \(\displaystyle 20{,}000\): \[n - 1 = 10 \;\Rightarrow\; n = 11.\]Reading the answer carefully. \(\displaystyle n = 11\) means the salary is ₹$\displaystyle 7,00,000$ in his \(\displaystyle 11^{\text{th}}\) year of work. He began the \(\displaystyle 1^{\text{st}}\) year on ₹$\displaystyle 5,00,000$, so he needed \(\displaystyle 10\) increments — that is, \(\displaystyle 10\) years of service — for the salary to climb to ₹$\displaystyle 7,00$,000.Check. \(\displaystyle 10\) increments add \(\displaystyle 10 \times 20{,}000 = 2{,}00{,}000\), and \(\displaystyle 5{,}00{,}000 + 2{,}00{,}000 = 7{,}00{,}000\). ✓Answer. After \(\displaystyle 10\) years — his salary is ₹$\displaystyle 7,00,000$ in the \(\displaystyle 11^{\text{th}}\) year.

Sum of an AP.Row $\displaystyle 1$ has \(\displaystyle 1\) marble, row $\displaystyle 2$ has \(\displaystyle 2\), row $\displaystyle 3$ has \(\displaystyle 3\), and so on up to row 25. The total number of marbles is therefore \[1 + 2 + 3 + \cdots + 25,\] which is the sum of an AP with \(\displaystyle a = 1\), \(\displaystyle d = 1\), \(\displaystyle n = 25\) and last term \(\displaystyle l = 25\).The pairing trick. Write the sum forwards and backwards, one under the other: \[\begin{aligned} S &= 1 + 2 + 3 + \cdots + 24 + 25 \\ S &= 25 + 24 + 23 + \cdots + 2 + 1 \end{aligned}\] Adding the two lines column by column, every column gives \(\displaystyle 26\), and there are \(\displaystyle 25\) columns: \[2S = 25 \times 26 = 650 \;\Rightarrow\; S = \frac{650}{2} = 325.\]This is the formula \(\displaystyle S_{n} = \dfrac{n}{2}(a + l) = \dfrac{25}{2}(1 + 25) = \dfrac{25 \times 26}{2} = 325\).Check on a small case. The same method for just \(\displaystyle 4\) rows gives \(\displaystyle \dfrac{4 \times 5}{2} = 10\), and indeed \(\displaystyle 1 + 2 + 3 + 4 = 10\). ✓ Adding row by row up to \(\displaystyle 25\) also gives \(\displaystyle 325\).Answer. The child uses \(\displaystyle 325\) marbles in all.