Class 9 Maths Ganita Manjari Chapter 8 Solutions — Predicting What Comes Next: Exploring Sequences and Progressions

Here $a=11$ and $d=-3$. Explicit rule: $t_{n}=11-3\left(n-1\right)=14-3n$, $n\ge 1$. Recursive rule: $t_{1}=11$; $t_{n}=t_{n-1}-3$ for $n\ge 2$. The initial term and the allowed indices are both part of a recursive definition.

The multiples are 12,15,…,99. Solve $12+3\left(n-1\right)=99$ to get $n=30$. For the sum, write the list forward and backward: each paired column is $12+99=111$, and there are 30 such columns. Thus $2S=30\times 111$ and $S=1665$. There are 30 numbers.

From $t_{1}=2$, successive use of $t_{n+1}=3t_{n}-2$ gives 2,4,10,28,82,244,730. Therefore 730 is the 7th term.

For an explicit derivation, subtract 1: $t_{n+1}-1=3\left(t_{n}-1\right)$. Thus $u_{n}=t_{n}-1$ is a GP with $u_{1}=1$ and ratio 3, so $t_{n}=3^{n-1}+1$. Setting this equal to 730 gives $3^{n-1}=729=3^{6}$, hence $n=7$.

With $a=2$,$r=3$, the explicit rule is $t_{n}=2\cdot 3^{n-1}$ for $n\ge 1$. The recursive rule is $t_{1}=2$; $t_{n}=3t_{n-1}$ for $n\ge 2$. Solve $2\cdot 3^{n-1}=4374$: $3^{n-1}=2187=3^{7}$, giving $n=8$.

Write $H_{k}=80\left(0.6\right)^{k}m$, where $k=0$ is the initial height and $k=1$ is the first rebound. (i) $H_{5}=80\left(0.6\right)^{5}=6.2208m$.

(ii) The first ground contact follows the initial 80 m fall. By the sixth ground contact there have been five complete up-and-down bounces, so $$D=80+2\left(H_{1}+H_{2}+H_{3}+H_{4}+H_{5}\right)\text{.}$$

The rebound heights are 48,28.8,17.28,10.368,6.2208 m. Their sum is 110.6688 m, hence $D=301.3376m\approx 301.34m$. Do not count a sixth rebound before the sixth ground contact.

(i) The numbers of smallest retained red squares at Stages 0,1,2,3 are 1,8,64,512. (ii) Stages 4,5 have 4096,32768. Each square produces 8 smaller retained squares.

(iii) If $N_{n}$ is the count at Stage n, then $N_{n}=8^{n}$ for $n\ge 0$; recursively $N_{0}=1$, $N_{n}=8N_{n-1}$ for $n\ge 1$.

(iv) Each retained small square has area $\frac{1}{9^{n}}$, so total area $A_{n}=\frac{8^{n}}{9^{n}}=\left(\frac{8}{9}\right)^{n}$. Stages 1–5 have areas $\frac{8}{9}$,$\frac{64}{81}$,$\frac{512}{729}$,$\frac{4096}{6561}$,$\frac{32768}{59049}$ square units. Recursively $A_{0}=1$; $A_{n}=\left(\frac{8}{9}\right)A_{n-1}$, $n\ge 1$. The area stays positive at every finite stage but tends to 0 as n increases without bound.

Let the first term be a and the ratio r. Then $a\left(1+r\right)=-4$ and $ar^{4}=4ar^{2}$. Since $a\ne 0$, $r^{2}\left(r^{2}-4\right)=0$.

Under the usual convention of nonzero terms and a defined consecutive-term ratio, $r\ne 0$, so $r=2$ or −2. If $r=2$, $a=-\frac{4}{3}$ and the GP is $-\frac{4}{3}$,$-\frac{8}{3}$,$-\frac{16}{3}$,$-\frac{32}{3}$,…. If $r=-2$, $a=4$ and the GP is 4,−8,16,−32,…. Both satisfy the conditions.

Convention note: if a geometric progression is allowed to mean any fixed-multiplier recurrence including $r=0$, the degenerate sequence −4,0,0,… also satisfies the equations. Its later consecutive quotients are undefined, so it is excluded by the nonzero-ratio convention.

Suppose k positive consecutive integers start at $m\ge 1$. Their sum is $\frac{k\left(2m+k-1\right)}{2}=100$, hence $m=\frac{\frac{200}{k}-k+1}{2}$. Since $1+\ldots +14=105$, $k\le 13$. Also k must divide 200. Candidates within that range are 1,2,4,5,8,10.

They give $m=100$,49.5,23.5,18,9,5.5 respectively. Only positive integers are allowed, so the complete list is 100 (one term), 18+19+20+21+22, and 9+10+11+12+13+14+15+16. If “sum” requires at least two terms, retain only the last two.

$S_{n}=\frac{n\left(n+1\right)}{2}$. Check the neighbouring values: $S_{44}=44\times \frac{45}{2}=990$, while $S_{45}=45\times \frac{46}{2}=1035$. Since these sums increase with n, the smallest n with $S_{n}>1000$ is 45.

Here $a=2$,$r=4$. Explicitly $t_{n}=2\cdot 4^{n-1}$ for $n\ge 1$. Recursively $t_{1}=2$; $t_{n}=4t_{n-1}$ for $n\ge 2$. Since $\frac{131072}{2}=65536=4^{8}$, $n-1=8$ and the requested value is the 9th term.

Write the three real, nonzero terms as $\frac{u}{r}$,u,ur. Their product is $u^{3}=-1$, so $u=-1$. The sum gives $-\frac{1}{r}-1-r=\frac{13}{12}$, or $r+\frac{1}{r}=-\frac{25}{12}$. Multiplying by 12r gives $$\begin{aligned}12r^{2}+25r+12&=0\\&=\left(3r+4\right)\left(4r+3\right)\text{.}\end{aligned}$$

Thus $r=-\frac{3}{4}$ with terms $\frac{4}{3}$,−1,$\frac{3}{4}$, or $r=-\frac{4}{3}$ with terms $\frac{3}{4}$,−1,$\frac{4}{3}$. In either order the sum is$\frac{16-12+9}{12}=\frac{13}{12}$ and the product is −1.

If the original first term is a and ratio r, then $x=ar^{3}$,$y=ar^{9}$,$z=ar^{15}$. Hence $y=x\cdot r^{6}$ and $z=y\cdot r^{6}$, so these three terms themselves form a GP with multiplier $r^{6}$. For nonzero terms, $\frac{y}{x}=\frac{z}{y}=r^{6}$. This directly proves the equal-ratio claim; it also gives $y^{2}=xz$.

Let the three terms be x,y,z, with $y^{2}=xz$. Given $x+y+z=26$ and $x^{2}+y^{2}+z^{2}=364$, squaring the first equation gives$676=364+2\left(xy+yz+xz\right)$, so $xy+yz+xz=156$.

Now $$\begin{aligned}xy+yz+xz&=y\left(x+z\right)+y^{2}\\&=y\left(x+y+z\right)\\&=26y\text{,}\end{aligned}$$ giving $y=6$. Hence $x+z=20$ and $xz=36$. The outer terms are roots of $t^{2}-20t+36=\left(t-2\right)\left(t-18\right)=0$. Thus the GP is 2,6,18 (ratio 3) or 18,6,2 (ratio$\frac{1}{3}$). Check: $4+36+324=364$.

Successive substitution gives 1,2,4,8,16,32,64,128. For $n\ge 4$, subtract $P_{n-1}=P_{1}+\ldots +P_{n-2}+1$ from the formula for $P_{n}$, obtaining $P_{n}-P_{n-1}=P_{n-1}$. The same doubling rule also holds directly at $n=2$,3.

A simpler complete recursive rule is $P_{1}=1$; $P_{n}=2P_{n-1}$ for $n\ge 2$. Its explicit form is $P_{n}=2^{n-1}$, $n\ge 1$.

$W_{3}=W_{1}+2=3$, $W_{4}=W_{1}+W_{2}+2=5$, and continuing gives 1,2,3,5,8,13,21,34. For $n\ge 4$, subtract the formula for $W_{n-1}$ from that for $W_{n}$: $W_{n}-W_{n-1}=W_{n-2}$. Since $W_{3}=W_{2}+W_{1}$ also holds, the recurrence is $W_{n}=W_{n-1}+W_{n-2}$ for $n\ge 3$, starting with 1,2. This is the textbook’s Virahānka–Fibonacci sequence.