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

Two terms are enough to find \(\displaystyle d\).In an AP the \(\displaystyle n^{\text{th}}\) term is \(\displaystyle a_n=a+(n-1)d\). To get from the \(\displaystyle 11^{\text{th}}\) term to the \(\displaystyle 16^{\text{th}}\) term you add \(\displaystyle d\) five times, so \[a_{16}-a_{11}=5d\quad\Longrightarrow\quad 73-38=5d\quad\Longrightarrow\quad 5d=35\quad\Longrightarrow\quad d=7.\]Now find the first term. \[a_{11}=a+10d=38\quad\Longrightarrow\quad a+70=38\quad\Longrightarrow\quad a=-32.\]The \(\displaystyle 31^{\text{st}}\) term. \[a_{31}=a+30d=-32+30(7)=-32+210=178.\]Check a second way. Go from the \(\displaystyle 16^{\text{th}}\) term instead, adding \(\displaystyle d\) fifteen times: \[a_{31}=a_{16}+15d=73+15(7)=73+105=178.\ \checkmark\]The \(\displaystyle 31^{\text{st}}\) term is 178.

Use the gap between two terms to get \(\displaystyle d\).Going from the \(\displaystyle 5^{\text{th}}\) term to the \(\displaystyle 7^{\text{th}}\) term adds \(\displaystyle d\) twice, so \[a_7-a_5=2d.\] We are told this gap is $\displaystyle 12$, so \[2d=12\quad\Longrightarrow\quad d=6.\]Then use the third term to get \(\displaystyle a\). \[a_3=a+2d=16\quad\Longrightarrow\quad a+12=16\quad\Longrightarrow\quad a=4.\]Write out the AP. With \(\displaystyle a=4\) and \(\displaystyle d=6\): \[4,\ 10,\ 16,\ 22,\ 28,\ 34,\ \ldots\] and in general \[a_n=4+(n-1)6=6n-2.\]Check. \(\displaystyle a_3=6(3)-2=16\ \checkmark\), and \(\displaystyle a_7-a_5=(6\cdot 7-2)-(6\cdot 5-2)=40-28=12\ \checkmark\).The AP is \(\displaystyle 4,\ 10,\ 16,\ 22,\ 28,\ \ldots\) (first term $\displaystyle 4$, common difference $\displaystyle 6$), with \(\displaystyle a_n=6n-2\).

Find the first and last three-digit multiples of 7.Three-digit numbers run from $\displaystyle 100$ to 999. These form an AP. The multiples of $\displaystyle 7$ in between go up in steps of $\displaystyle 7$: \[105,\ 112,\ 119,\ \ldots,\ 994,\qquad a=105,\ d=7.\]Count the terms. Let $\displaystyle 994$ be the \(\displaystyle n^{\text{th}}\) term: \[a+(n-1)d=994\] \[105+(n-1)7=994\] \[(n-1)7=889\] \[n-1=127\quad\Longrightarrow\quad n=128.\]Check another way. The multiples are \(\displaystyle 7\times 15,\ 7\times 16,\ \ldots,\ 7\times 142\), so we are counting the whole numbers from $\displaystyle 15$ to $\displaystyle 142$, and there are \(\displaystyle 142-15+1=128\) of them. \(\displaystyle \checkmark\)There are $\displaystyle 128$ three-digit numbers divisible by 7.

Find the first and last multiples of $\displaystyle 4$ in the range. (Neither $\displaystyle 10$ nor $\displaystyle 250$ is itself a multiple of $\displaystyle 4$, so it makes no difference here whether "between" includes the end numbers.)These form an AP. \[12,\ 16,\ 20,\ \ldots,\ 248,\qquad a=12,\ d=4.\]Count the terms. Let $\displaystyle 248$ be the \(\displaystyle n^{\text{th}}\) term: \[12+(n-1)4=248\] \[(n-1)4=236\] \[n-1=59\quad\Longrightarrow\quad n=60.\]Check another way. The numbers are \(\displaystyle 4\times 3,\ 4\times 4,\ \ldots,\ 4\times 62\), that is the whole numbers from $\displaystyle 3$ to $\displaystyle 62$, and there are \(\displaystyle 62-3+1=60\) of them. \(\displaystyle \checkmark\)$\displaystyle 60$ multiples of $\displaystyle 4$ lie between $\displaystyle 10$ and 250.

Let the GP be \(\displaystyle a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4},\ \ldots\) In a GP neither \(\displaystyle a\) nor \(\displaystyle r\) can be $\displaystyle 0$, which is what lets us divide by them below.Use the second condition first — it gives \(\displaystyle r\) on its own. "The fifth term is $\displaystyle 4$ times the third term" means \[ar^{4}=4\,ar^{2}.\] Divide both sides by \(\displaystyle ar^{2}\) (allowed, since \(\displaystyle a\neq 0\) and \(\displaystyle r\neq 0\)): \[r^{2}=4\quad\Longrightarrow\quad r=2\ \text{ or }\ r=-2.\] Both signs must be kept — squaring loses the sign, so we test each.Now use the first condition. "The sum of the first two terms is \(\displaystyle -4\)" means \[a+ar=a(1+r)=-4.\]Case \(\displaystyle r=2\): \(\displaystyle a(1+2)=-4\Rightarrow 3a=-4\Rightarrow a=-\dfrac{4}{3}\). The GP is \[-\frac{4}{3},\ -\frac{8}{3},\ -\frac{16}{3},\ -\frac{32}{3},\ -\frac{64}{3},\ \ldots\] Check: \(\displaystyle -\tfrac43-\tfrac83=-\tfrac{12}{3}=-4\ \checkmark\); fifth term \(\displaystyle -\tfrac{64}{3}=4\times\left(-\tfrac{16}{3}\right)\ \checkmark\).Case \(\displaystyle r=-2\): \(\displaystyle a(1-2)=-4\Rightarrow -a=-4\Rightarrow a=4\). The GP is \[4,\ -8,\ 16,\ -32,\ 64,\ \ldots\] Check: \(\displaystyle 4+(-8)=-4\ \checkmark\); fifth term \(\displaystyle 64=4\times 16\ \checkmark\).Two GPs satisfy both conditions: \(\displaystyle 4,\ -8,\ 16,\ -32,\ 64,\ \ldots\) (with \(\displaystyle a=4,\ r=-2\)) and \(\displaystyle -\dfrac{4}{3},\ -\dfrac{8}{3},\ -\dfrac{16}{3},\ -\dfrac{32}{3},\ \ldots\) (with \(\displaystyle a=-\dfrac{4}{3},\ r=2\)). Either one is a correct answer to "find a GP".

Set it up as an AP with common difference 1.Suppose $\displaystyle 100$ is the sum of \(\displaystyle k\) consecutive natural numbers starting at \(\displaystyle a\): \[a+(a+1)+(a+2)+\cdots+(a+k-1)=100,\qquad a\ge 1,\ k\ge 1.\] This is an AP with first term \(\displaystyle a\), common difference $\displaystyle 1$ and \(\displaystyle k\) terms, so its sum is \[\frac{k}{2}\Big[2a+(k-1)\Big]=100\quad\Longrightarrow\quad \boxed{k(2a+k-1)=200}.\]Two facts cut the search down to almost nothing.Fact $\displaystyle 1$ — one factor must be odd. The numbers \(\displaystyle k\) and \(\displaystyle 2a+k-1\) differ by \(\displaystyle 2a-1\), an odd number, so one of them is even and the other is odd. Since \(\displaystyle 200=2^{3}\times 5^{2}\), the only odd divisors of $\displaystyle 200$ are \(\displaystyle 1,\ 5,\ 25\). So one of \(\displaystyle k\) and \(\displaystyle 2a+k-1\) is $\displaystyle 1$, $\displaystyle 5$ or 25.Fact $\displaystyle 2$ — \(\displaystyle k\) is the smaller factor. Because \(\displaystyle a\ge 1\), \[2a+k-1\ \ge\ k+1\ >\ k,\] so \(\displaystyle k\cdot k<k(2a+k-1)=200\), giving \(\displaystyle k^{2}<200\) and therefore \(\displaystyle k\le 14\).Check each possibility.If \(\displaystyle k\) is the odd one: \(\displaystyle k\in\{1,5,25\}\), and \(\displaystyle k\le 14\) rules out 25. If \(\displaystyle 2a+k-1\) is the odd one: \(\displaystyle 2a+k-1\in\{1,5,25\}\), so \(\displaystyle k=200,\ 40\) or \(\displaystyle 8\); only \(\displaystyle k=8\) satisfies \(\displaystyle k\le 14\). Verify both sums. \[18+19+20+21+22=100\ \checkmark\qquad\text{(five terms averaging }20)\] \[9+10+11+12+13+14+15+16=\frac{8(9+16)}{2}=4\times 25=100\ \checkmark\]A quick computer search over all starting values also finds exactly these two (plus the one-term case), which confirms nothing was missed.There are exactly two ways to write $\displaystyle 100$ as a sum of two or more consecutive natural numbers: \[100=18+19+20+21+22\] \[100=9+10+11+12+13+14+15+16\] (If a "sum" of a single number is allowed, \(\displaystyle 100=100\) is a third, trivial, case.)

Doubling means a GP with \(\displaystyle r=2\).Start with $\displaystyle 30$ bacteria. Each hour the count is multiplied by $\displaystyle 2$, so the counts form a GP: \[\underbrace{30}_{\text{start}},\ \underbrace{60}_{\text{1st hour}},\ \underbrace{120}_{\text{2nd hour}},\ \underbrace{240}_{\text{3rd hour}},\ \underbrace{480}_{\text{4th hour}},\ \ldots\]Notice the count at the end of the \(\displaystyle n^{\text{th}}\) hour has been doubled \(\displaystyle n\) times, so the exponent is \(\displaystyle n\) itself (not \(\displaystyle n-1\)) — the starting count $\displaystyle 30$ belongs to hour $\displaystyle 0$, before any doubling.End of the \(\displaystyle 2^{\text{nd}}\) hour. \[30\times 2^{2}=30\times 4=120.\]End of the \(\displaystyle 4^{\text{th}}\) hour. \[30\times 2^{4}=30\times 16=480.\]End of the \(\displaystyle n^{\text{th}}\) hour. \[30\times 2^{\,n}.\]Check. Put \(\displaystyle n=2\): \(\displaystyle 30\times 4=120\ \checkmark\). Put \(\displaystyle n=0\): \(\displaystyle 30\times 1=30\), the original count \(\displaystyle \checkmark\).$\displaystyle 120$ bacteria after $\displaystyle 2$ hours, $\displaystyle 480$ after $\displaystyle 4$ hours, and \(\displaystyle 30\times 2^{\,n}\) after \(\displaystyle n\) hours.

Turn each statement into an equation in \(\displaystyle a\) and \(\displaystyle d\).With \(\displaystyle a_n=a+(n-1)d\): \[a_4+a_8=(a+3d)+(a+7d)=2a+10d=24\quad\Longrightarrow\quad a+5d=12 \tag{1}\] \[a_6+a_{10}=(a+5d)+(a+9d)=2a+14d=44\quad\Longrightarrow\quad a+7d=22 \tag{2}\] (Each equation was halved, which keeps the numbers small.)Subtract to eliminate \(\displaystyle a\). Doing \(\displaystyle (2)-(1)\): \[(a+7d)-(a+5d)=22-12\] \[2d=10\quad\Longrightarrow\quad d=5.\]Back-substitute into ($\displaystyle 1$). \[a+5(5)=12\quad\Longrightarrow\quad a+25=12\quad\Longrightarrow\quad a=-13.\]First three terms. \[a=-13,\qquad a+d=-8,\qquad a+2d=-3.\]Check. \(\displaystyle a_4=-13+3(5)=2\) and \(\displaystyle a_8=-13+7(5)=22\), so \(\displaystyle a_4+a_8=24\ \checkmark\). Also \(\displaystyle a_6=12\) and \(\displaystyle a_{10}=32\), so \(\displaystyle a_6+a_{10}=44\ \checkmark\).The first three terms are \(\displaystyle -13,\ -8,\ -3\) (with \(\displaystyle a=-13\) and \(\displaystyle d=5\)).

Use the formula for the sum of the first \(\displaystyle n\) natural numbers. \[1+2+3+\cdots+n=\frac{n(n+1)}{2}.\] We need the smallest \(\displaystyle n\) with \[\frac{n(n+1)}{2}>1000\quad\Longrightarrow\quad n(n+1)>2000.\]Estimate first, then test. Since \(\displaystyle n(n+1)\) is a little more than \(\displaystyle n^{2}\), and \(\displaystyle \sqrt{2000}\approx 44.7\), the answer must be near $\displaystyle 44$ or 45. So test both rather than guessing: \[n=44:\quad \frac{44\times 45}{2}=\frac{1980}{2}=990,\qquad 990\not>1000.\] \[n=45:\quad \frac{45\times 46}{2}=\frac{2070}{2}=1035,\qquad 1035>1000.\ \checkmark\]Why no smaller \(\displaystyle n\) can work. The sums \(\displaystyle \frac{n(n+1)}{2}\) increase as \(\displaystyle n\) increases (each step adds the positive number \(\displaystyle n+1\)). So if \(\displaystyle n=44\) already falls short, every \(\displaystyle n\le 44\) falls short too. Therefore $\displaystyle 45$ is the smallest.The smallest value is \(\displaystyle n=45\), and then \(\displaystyle 1+2+\cdots+45=1035>1000\).

Find the common ratio. \[\frac{8}{2}=4,\qquad \frac{32}{8}=4,\] so this is a GP with \(\displaystyle a=2\) and \(\displaystyle r=4\), and \[a_n=2\cdot 4^{\,n-1}.\]Compare powers of 2. Both $\displaystyle 2$ and $\displaystyle 4$ are powers of $\displaystyle 2$, and so is $\displaystyle 131072$, so rewrite everything with base 2. Since \(\displaystyle 4=2^{2}\), \[a_n=2^{1}\cdot\left(2^{2}\right)^{n-1}=2^{1+2(n-1)}=2^{\,2n-1}.\] Now factor $\displaystyle 131072$ as a power of $\displaystyle 2$ by halving repeatedly: \[131072=2^{17}\qquad(\text{because }2^{10}=1024\text{ and }2^{7}=128,\ \text{and }1024\times 128=131072).\]Equate the exponents. \[2^{\,2n-1}=2^{17}\quad\Longrightarrow\quad 2n-1=17\quad\Longrightarrow\quad 2n=18\quad\Longrightarrow\quad n=9.\]Check. \(\displaystyle a_9=2\cdot 4^{8}=2\times 65536=131072\ \checkmark\).Formulas. \[\textbf{Explicit: }\ a_n=2\cdot 4^{\,n-1}\ \left(=2^{\,2n-1}\right),\qquad\qquad \textbf{Recursive: }\ a_1=2,\ \ a_n=4\,a_{n-1}\ \ (n\ge 2).\]$\displaystyle 131072$ is the \(\displaystyle 9^{\text{th}}\) term; \(\displaystyle a_n=2\cdot 4^{\,n-1}\), and \(\displaystyle a_1=2,\ a_n=4a_{n-1}\).