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

Choose the terms symmetrically. For three terms of a GP it is far easier to write them as \[\frac{a}{r},\qquad a,\qquad ar\] than as \(\displaystyle a,\ ar,\ ar^{2}\) — because then the product collapses at once.Use the product. \[\frac{a}{r}\cdot a\cdot ar=a^{3}=-1\quad\Longrightarrow\quad a=-1,\] since \(\displaystyle -1\) is the only real number whose cube is \(\displaystyle -1\).Use the sum. With \(\displaystyle a=-1\), \[\frac{-1}{r}+(-1)+(-1)r=\frac{13}{12}\] \[-\left(\frac{1}{r}+1+r\right)=\frac{13}{12}\] \[\frac{1}{r}+r=-\frac{13}{12}-1=-\frac{25}{12}.\] Multiply through by \(\displaystyle 12r\) (valid because \(\displaystyle r\neq 0\) in a GP): \[12+12r^{2}=-25r\quad\Longrightarrow\quad 12r^{2}+25r+12=0.\]Factorise. Split the middle term using \(\displaystyle 9+16=25\) and \(\displaystyle 9\times 16=144=12\times 12\): \[12r^{2}+9r+16r+12=3r(4r+3)+4(4r+3)=(3r+4)(4r+3)=0,\] so \[r=-\frac{4}{3}\qquad\text{or}\qquad r=-\frac{3}{4}.\]Write out the terms. With \(\displaystyle a=-1\) and \(\displaystyle r=-\dfrac{3}{4}\): \[\frac{a}{r}=\frac{-1}{-3/4}=\frac{4}{3},\qquad a=-1,\qquad ar=(-1)\left(-\frac{3}{4}\right)=\frac{3}{4}.\] So the terms are \(\displaystyle \dfrac{4}{3},\ -1,\ \dfrac{3}{4}\). The other root \(\displaystyle r=-\dfrac{4}{3}\) gives the very same three numbers in the reverse order, \(\displaystyle \dfrac{3}{4},\ -1,\ \dfrac{4}{3}\).Check. \[\frac{4}{3}+(-1)+\frac{3}{4}=\frac{16-12+9}{12}=\frac{13}{12}\ \checkmark\qquad \frac{4}{3}\times(-1)\times\frac{3}{4}=-1\ \checkmark\]Common ratio \(\displaystyle r=-\dfrac{3}{4}\) (or \(\displaystyle r=-\dfrac{4}{3}\), which just reverses the order), and the three terms are \(\displaystyle \dfrac{4}{3},\ -1,\ \dfrac{3}{4}\).

Write all three terms from the same first term and ratio.Let the GP have first term \(\displaystyle A\) and common ratio \(\displaystyle R\), so its \(\displaystyle n^{\text{th}}\) term is \(\displaystyle A R^{\,n-1}\). (In a GP \(\displaystyle A\neq 0\) and \(\displaystyle R\neq 0\).) Then \[x=a_4=AR^{3},\qquad y=a_{10}=AR^{9},\qquad z=a_{16}=AR^{15}.\]Method $\displaystyle 1$ — show the two ratios are equal. Three numbers are in GP exactly when the second divided by the first equals the third divided by the second. Here \[\frac{y}{x}=\frac{AR^{9}}{AR^{3}}=R^{\,9-3}=R^{6},\qquad\quad \frac{z}{y}=\frac{AR^{15}}{AR^{9}}=R^{\,15-9}=R^{6}.\] The two ratios are both \(\displaystyle R^{6}\), so they are equal. Hence \(\displaystyle x,\ y,\ z\) form a GP with common ratio \(\displaystyle R^{6}\).Method $\displaystyle 2$ — show \(\displaystyle y^{2}=xz\). This is the same statement rearranged, and needs no division: \[y^{2}=\left(AR^{9}\right)^{2}=A^{2}R^{18},\] \[xz=\left(AR^{3}\right)\left(AR^{15}\right)=A^{2}R^{\,3+15}=A^{2}R^{18}.\] So \(\displaystyle y^{2}=xz\), which is precisely the condition for \(\displaystyle x,\ y,\ z\) to be in GP.Why it works. The term numbers $\displaystyle 4$, $\displaystyle 10$, $\displaystyle 16$ are themselves equally spaced (each $\displaystyle 6$ apart). Stepping $\displaystyle 6$ places along a GP always multiplies by \(\displaystyle R^{6}\), so picking terms at equal gaps out of a GP always leaves you with another GP.Hence \(\displaystyle x,\ y,\ z\) are in GP, with common ratio \(\displaystyle R^{6}\). Proved.

Set up both conditions. Let the three terms be \(\displaystyle a,\ ar,\ ar^{2}\). Then \[a+ar+ar^{2}=a\left(1+r+r^{2}\right)=26 \tag{1}\] \[a^{2}+a^{2}r^{2}+a^{2}r^{4}=a^{2}\left(1+r^{2}+r^{4}\right)=364 \tag{2}\]The key identity. The awkward factor in ($\displaystyle 2$) factorises: \[1+r^{2}+r^{4}=\left(1+r+r^{2}\right)\left(1-r+r^{2}\right).\] Expanding the right side confirms it: \(\displaystyle 1-r+r^{2}+r-r^{2}+r^{3}+r^{2}-r^{3}+r^{4}=1+r^{2}+r^{4}\ \checkmark\)Divide ($\displaystyle 2$) by the square of ($\displaystyle 1$). Squaring ($\displaystyle 1$) gives \(\displaystyle a^{2}\left(1+r+r^{2}\right)^{2}=676\). Dividing ($\displaystyle 2$) by that: \[\frac{a^{2}\left(1+r+r^{2}\right)\left(1-r+r^{2}\right)}{a^{2}\left(1+r+r^{2}\right)^{2}}=\frac{364}{676}\] \[\frac{1-r+r^{2}}{1+r+r^{2}}=\frac{364}{676}=\frac{7}{13}.\] The whole point of dividing is that \(\displaystyle a\) cancels, leaving an equation in \(\displaystyle r\) alone. (Cancelling \(\displaystyle 1+r+r^{2}\) is safe: for every real \(\displaystyle r\) it equals \(\displaystyle \left(r+\tfrac12\right)^{2}+\tfrac34>0\), so it is never zero.)Solve for \(\displaystyle r\). Cross-multiplying: \[13\left(1-r+r^{2}\right)=7\left(1+r+r^{2}\right)\] \[13-13r+13r^{2}=7+7r+7r^{2}\] \[6r^{2}-20r+6=0\quad\Longrightarrow\quad 3r^{2}-10r+3=0.\] Split the middle term: \(\displaystyle 3r^{2}-9r-r+3=3r(r-3)-1(r-3)=(3r-1)(r-3)=0\), so \[r=3\qquad\text{or}\qquad r=\frac{1}{3}.\]Find \(\displaystyle a\) for each root, using ($\displaystyle 1$). Check. \[2+6+18=26\ \checkmark\qquad 2^{2}+6^{2}+18^{2}=4+36+324=364\ \checkmark\]The terms are \(\displaystyle 2,\ 6,\ 18\) (or \(\displaystyle 18,\ 6,\ 2\), the same GP taken in the reverse order).

Compute the terms one at a time, straight from the rule. The rule for \(\displaystyle n>2\) is: add up all the earlier terms, then add 1. \[P_1=1,\qquad P_2=2\] \[P_3=P_1+P_2+1=1+2+1=4\] \[P_4=P_1+P_2+P_3+1=1+2+4+1=8\] \[P_5=1+2+4+8+1=16\] \[P_6=1+2+4+8+16+1=32\] \[P_7=1+2+4+8+16+32+1=64\] \[P_8=1+2+4+8+16+32+64+1=128\]

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle P_n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 4$	$\displaystyle 8$	$\displaystyle 16$	$\displaystyle 32$	$\displaystyle 64$	$\displaystyle 128$

A simpler recursive formula. Each term is clearly double the one before it. Here is why, without checking case by case. For \(\displaystyle n>3\), write the rule twice: \[P_n=\left(P_1+P_2+\cdots+P_{n-2}+P_{n-1}\right)+1,\] \[P_{n-1}=\left(P_1+P_2+\cdots+P_{n-2}\right)+1.\] Subtracting the second from the first, everything inside the brackets cancels except \(\displaystyle P_{n-1}\): \[P_n-P_{n-1}=P_{n-1}\quad\Longrightarrow\quad P_n=2P_{n-1}.\] And \(\displaystyle P_3=4=2\times 2=2P_2\) directly, while \(\displaystyle P_2=2=2\times 1=2P_1\) as well. So \[\textbf{Simpler recursive formula: }\ P_1=1,\qquad P_n=2P_{n-1}\ \ \text{for all } n\ge 2.\] This says the sequence is just a GP with first term $\displaystyle 1$ and common ratio $\displaystyle 2$ — the "sum of everything before, plus $\displaystyle 1$" rule was a disguised doubling.Explicit formula. Starting at $\displaystyle 1$ and doubling \(\displaystyle n-1\) times, \[\textbf{Explicit: }\ P_n=2^{\,n-1}.\]Check it against the original rule. For \(\displaystyle n>2\), the earlier terms are \(\displaystyle 1+2+4+\cdots+2^{\,n-2}\), a GP sum equal to \(\displaystyle 2^{\,n-1}-1\). Adding $\displaystyle 1$ gives \(\displaystyle 2^{\,n-1}=P_n\ \checkmark\). (For example \(\displaystyle P_8=2^{7}=128\), matching the table.)\(\displaystyle P_1,\ldots,P_8\) are \(\displaystyle 1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ 128\); the simpler recursion is \(\displaystyle P_n=2P_{n-1}\) with \(\displaystyle P_1=1\); and the explicit formula is \(\displaystyle P_n=2^{\,n-1}\).

Read the rule carefully — it stops two terms short. For \(\displaystyle n>2\), \(\displaystyle W_n\) adds up the terms only as far as \(\displaystyle W_{n-2}\) (not \(\displaystyle W_{n-1}\)), and then adds 2. \[W_1=1,\qquad W_2=2\] \[W_3=W_1+2=1+2=3\] \[W_4=W_1+W_2+2=1+2+2=5\] \[W_5=W_1+W_2+W_3+2=1+2+3+2=8\] \[W_6=1+2+3+5+2=13\] \[W_7=1+2+3+5+8+2=21\] \[W_8=1+2+3+5+8+13+2=34\]

\(\displaystyle n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle W_n\)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 5$	$\displaystyle 8$	$\displaystyle 13$	$\displaystyle 21$	$\displaystyle 34$

Do you recognise it? Yes — \(\displaystyle 1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ 34\) are the Fibonacci numbers: from the third term on, each one is the sum of the two before it \(\displaystyle (1+2=3,\ 2+3=5,\ 3+5=8,\ 5+8=13,\ 8+13=21,\ 13+21=34)\).Why the rule produces Fibonacci. For \(\displaystyle n>3\), write the rule twice: \[W_n=\left(W_1+\cdots+W_{n-3}+W_{n-2}\right)+2,\] \[W_{n-1}=\left(W_1+\cdots+W_{n-3}\right)+2.\] Subtracting, all the common terms cancel and only \(\displaystyle W_{n-2}\) survives: \[W_n-W_{n-1}=W_{n-2}\quad\Longrightarrow\quad W_n=W_{n-1}+W_{n-2}.\] That is exactly the Fibonacci rule. It also holds at \(\displaystyle n=3\), since \(\displaystyle W_3=3=2+1=W_2+W_1\).If the Fibonacci numbers are written \(\displaystyle F_1=F_2=1,\ F_3=2,\ F_4=3,\ F_5=5,\ \ldots\), then our sequence is that list shifted by one place: \[W_n=F_{n+1}.\]\(\displaystyle W_1,\ldots,W_8\) are \(\displaystyle 1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ 34\) — the Fibonacci sequence, obeying \(\displaystyle W_n=W_{n-1}+W_{n-2}\) (equivalently \(\displaystyle W_n=F_{n+1}\)).