Predicting What Comes Next : Exploring Sequences and Progressions — NCERT Solutions

• Natural numbers: each term is 1 more than the previous term, so the next few are 7, 8, 9, ...
• Odd numbers: each term increases by 2, so the next few are 13, 15, 17, ...
• Triangular numbers: the differences increase by 1 each time; the next few are 28, 36, 45, ...
• Square numbers: each term is the square of a natural number; the next few are 49, 64, 81, ...

Yes. Examples of finite sequences in daily life include:

• a number of people in each row of seats in a classroom
• pages in a chapter
• floors in a building
• scores in a game over a fixed number of rounds
• steps in a staircase with a fixed number of steps

Any such list with a fixed number of terms is a finite sequence.

An explicit formula is useful because it lets us find any term directly by substituting the position number $n$.

So we can find very far terms like the 20th, 53rd, or 300th term without calculating all the earlier terms first. We can also check whether a number belongs to the sequence and find its position.

The sequence of square numbers is
$$1,4,9,16,25,\dots$$
These are the squares of natural numbers:
$$1=1^2,\;4=2^2,\;9=3^2,\;16=4^2,\;25=5^2,\dots$$
So the rule for the $n^{\text{th}}$ term is
$$\boxed{t_n=n^2}.$$

The chapter says prime numbers do not show a clear regular pattern like APs or GPs. So there is no simple rule given here that can predict the next few prime numbers exactly.

The prime sequence begins:
$$2,3,5,7,11,13,17,19,23,29,\dots$$
But unlike APs or GPs, there is no easy formula in this chapter for generating the next prime.

The sequence for the number of squares is a GP with first term $3$ and common ratio $2$.

So:

• Stage 5: $3\times 2^{5-1}=3\times 16=48$
• Stage 6: $3\times 2^{6-1}=3\times 32=96$
• Stage 10: $3\times 2^9=3\times 512=1536$
• Stage 11: $3\times 2^{10}=3072$
• Stage 12: $3\times 2^{11}=6144$
• Stage 20: $3\times 2^{19}=1,572,864$

In general, at stage $n$, the number of squares is
$$\boxed{3\times 2^{n-1}}.$$

The arithmetic progressions (APs) discussed in the chapter are the sequences in which the difference between consecutive terms is constant.

Examples of APs from the chapter:

• $1,4,7,10,\dots$
• $1,5,9,13,\dots$
• $11,7,3,-1,-5,\dots$
• $2,5,8,11,\dots$
• $-5,-1,3,7,\dots$

Not APs:

• natural numbers are an AP, yes, with difference 1
• odd numbers are an AP, yes, with difference 2
• triangular numbers are not an AP
• square numbers are not an AP
• prime numbers are not an AP
• geometric progressions are not APs because their ratio, not difference, is constant

So, only those sequences with a constant difference are APs.

Yes. The same method can be used.

Let
$$S=1+2+3+\cdots+100.$$
Reversing it,
$$S=100+99+98+\cdots+1.$$
Adding gives
$$2S=(1+100)+(2+99)+\cdots=(101)\text{ added 100 times}.$$
So
$$2S=100\times 101$$
and therefore
$$S=\frac{100\times 101}{2}=5050.$$

Using
$$S_n=\frac{n(n+1)}{2},$$
we get:

• $$S_{20}=\frac{20\times 21}{2}=10\times 21=210$$
• $$S_{50}=\frac{50\times 51}{2}=25\times 51=1275$$
• $$S_{1000}=\frac{1000\times 1001}{2}=500\times 1001=500500$$

The $n^{\text{th}}$ triangular number is
$$t_n=\frac{n(n+1)}{2}.$$
So:

• $$t_{10}=\frac{10\times 11}{2}=55$$
• $$t_{17}=\frac{17\times 18}{2}=153$$
• $$t_{80}=\frac{80\times 81}{2}=40\times 81=3240$$

This is a geometric progression. The number of squares doubles at each stage, so the sequence is:
$$3,6,12,24,48,96,\dots$$
Thus:

• Stage 5: 48
• Stage 6: 96
• Stage 10: 1536
• Stage 11: 3072
• Stage 12: 6144
• Stage 20: 1,572,864

In contrast to Fig. 8.3, where each stage increases by a fixed difference of 4 and forms an AP, this pattern has a fixed ratio of 2 and forms a GP.

No simple rule is given in the chapter for predicting prime numbers. The chapter only lists the first ten primes and notes that they do not show a clear regular pattern like APs or GPs.

The GP is
$$3,30,300,3000,\dots$$
Each term is obtained by multiplying the previous term by $10$.
So the recursive rule is:
$$\boxed{t_1=3,\; t_n=10t_{n-1}\text{ for }n\ge 2.}$$

The number of black triangles at Stages 0 to 3 is:

• Stage 0: 1
• Stage 1: 3
• Stage 2: 9
• Stage 3: 27

The number of black triangles triples at each stage, so:

• Stage 4: $27\times 3=81$
• Stage 5: $81\times 3=243$

The number of black triangles follows the GP
$$1,3,9,27,81,243,\dots$$
which is powers of 3.
So the rule for the $n^{\text{th}}$ stage is
$$\boxed{t_n=3^n}.$$

If the black region in Stage 0 has area 1 square unit, then:

• Stage 1: $\frac{3}{4}$
• Stage 2: $\left(\frac{3}{4}\right)^2$
• Stage 3: $\left(\frac{3}{4}\right)^3$
• Stage 4: $\left(\frac{3}{4}\right)^4$
• Stage 5: $\left(\frac{3}{4}\right)^5$

In general, the area of the black region at Stage $n$ is
$$\left(\frac{3}{4}\right)^n.$$
As $n$ increases, this area gets smaller and smaller and tends to 0.

Five different kinds of sequences could be:

1. Natural number sequence: $1,2,3,4,\dots$
2. Odd number sequence: $1,3,5,7,\dots$
3. Even number sequence: $2,4,6,8,\dots$
4. Square number sequence: $1,4,9,16,\dots$
5. Geometric sequence: $3,6,12,24,\dots$

Their properties differ: some increase by a fixed difference, some by a fixed ratio, and some follow special patterns like squares or triangular numbers.

The sequence is
$$1,4,7,10,13,\dots$$
Each term increases by 3, so the next four terms are
$$16,19,22,25.$$

The sequence of sums up to each term is:

• $1$
• $1+4=5$
• $1+4+7=12$
• $1+4+7+10=22$
• $1+4+7+10+13=35$
• $1+4+7+10+13+16=51$
• $\dots$

Thus the first ten terms of the sequence of running totals are:
$$1,5,12,22,35,51,70,92,117,145.$$

The triangular numbers are
$$1,3,6,10,15,21,28,36,\dots$$
So:

• $t_5=15$
• $t_6=21$
• $t_7=28$
• $t_8=36$

The sequence is given by
$$t_n=3n-7.$$
Now substitute the required values of $n$:

• $t_1=3(1)-7=-4$
• $t_2=6-7=-1$
• $t_3=9-7=2$
• $t_{12}=36-7=29$
• $t_{18}=54-7=47$
• $t_{50}=150-7=143$

So the required terms are
$$-4,-1,2,29,47,143.$$

These are not among the listed options because the question asks for a computed answer from the given rule.

We need to find $n$ such that
$$3n-7=332.$$
So,
$$3n=339$$
$$n=113.$$
Thus 332 is the 113th term.

Check whether 557 is a term by solving
$$3n-7=557.$$
Then
$$3n=564$$
$$n=188.$$
Since $188$ is a natural number, 557 is a term of the sequence. It is the 188th term.

Yes, both sequences are arithmetic progressions because the difference between consecutive terms is constant.

• For $2,5,8,11,\dots$, the common difference is $3$, so

$$t_n=2+(n-1)\cdot 3=3n-1.$$

• For $-5,-1,3,7,\dots$, the common difference is $4$, so

$$t_n=-5+(n-1)\cdot 4=4n-9.$$

When the ordered pairs are plotted, they lie on a straight line.

For an arithmetic progression (AP) with first term $a$ and common difference $d$, the $n^{\text{th}}$ term is

$$t_n = a + (n - 1)d.$$

For an AP, each term is obtained by adding the common difference $d$ to the previous term. So the recursive rule is

$$t_1 = a, \qquad t_n = t_{n-1} + d \text{ for } n \geq 2.$$

Substitute $n=1,2,3,4,5$.

(i) $t_n=3n-4$

• $t_1=3(1)-4=-1$
• $t_2=3(2)-4=2$
• $t_3=3(3)-4=5$
• $t_4=3(4)-4=8$
• $t_5=3(5)-4=11$

So the first five terms are $-1, 2, 5, 8, 11$.

(ii) $t_n=2-5n$

• $t_1=2-5=-3$
• $t_2=2-10=-8$
• $t_3=2-15=-13$
• $t_4=2-20=-18$
• $t_5=2-25=-23$

So the first five terms are $-3, -8, -13, -18, -23$.

(iii) $t_n=n^2-2n+3$

• $t_1=1-2+3=2$
• $t_2=4-4+3=3$
• $t_3=9-6+3=6$
• $t_4=16-8+3=11$
• $t_5=25-10+3=18$

So the first five terms are $2, 3, 6, 11, 18$.

Use $t_n=5n-3$.

• $t_{10}=5(10)-3=50-3=47$
• $t_{15}=5(15)-3=75-3=72$

So the required terms are 47 and 72.

For a number to be a term of the sequence, it must satisfy $t_n=5n-3$ for a natural number $n$.

For 97:
$$5n-3=97 \Rightarrow 5n=100 \Rightarrow n=20$$
Since $n=20$ is a natural number, 97 is a term.

For 172:
$$5n-3=172 \Rightarrow 5n=175 \Rightarrow n=35$$
Since $n=35$ is a natural number, actually 172 is also a term.

So both numbers are terms of the sequence.

Set $t_n=607$ in $t_n=5n-3$.

$$5n-3=607$$
$$5n=610$$
$$n=122$$

So 607 is the 122nd term.

The recursive rule is

$$t_1=-5,\qquad t_{n+1}=t_n+3.$$

So the terms are found by adding 3 each time:

• $t_1=-5$
• $t_2=-5+3=-2$
• $t_3=-2+3=1$
• $t_4=1+3=4$
• $t_5=4+3=7$

Hence the first five terms are $-5,-2,1,4,7$.

To check whether 52 is a term, write the AP formula:

$$t_n=-5+(n-1)\cdot 3=3n-8.$$

Set $t_n=52$:

$$3n-8=52$$
$$3n=60$$
$$n=20$$

So 52 is a term, and it is the 20th term.

Given

$$T_1=1,\ T_2=2,\ T_3=4,$$

and

$$T_n=T_{n-1}+T_{n-2}+T_{n-3} \quad (n\ge 4).$$

Now compute successively:

$$T_4=T_3+T_2+T_1=4+2+1=7$$
$$T_5=T_4+T_3+T_2=7+4+2=13$$
$$T_6=T_5+T_4+T_3=13+7+4=24$$
$$T_7=T_6+T_5+T_4=24+13+7=44$$
$$T_8=T_7+T_6+T_5=44+24+13=81$$

So the values are 7, 13, 24, 44, 81.

For the AP $3,8,13,18,\dots$, the first term is $a=3$ and common difference is $d=5$.

Use $t_n=a+(n-1)d$.

• $t_{10}=3+(10-1)\cdot 5=3+45=48$
• $t_{26}=3+(26-1)\cdot 5=3+125=128$

So the required terms are 48 and 128.

For the AP $21,18,15,\dots$, we have first term $a=21$ and common difference $d=-3$.

The $n^{\text{th}}$ term is

$$t_n=a+(n-1)d=21+(n-1)(-3)=24-3n.$$

To find which term is $-81$:

$$24-3n=-81$$
$$-3n=-105$$
$$n=35$$

So $-81$ is the 35th term.

Now check whether 0 is a term:

$$24-3n=0$$
$$3n=24$$
$$n=8$$

Since $n=8$ is a natural number, 0 is also a term of the AP.

For the AP $11,8,5,2,\dots$, the first term is $a=11$ and common difference is $d=-3$.

So

$$t_n=a+(n-1)d=11+(n-1)(-3)=11-3n+3=14-3n.$$

Thus the $n^{\text{th}}$ term is $t_n=14-3n$.

For the recursive rule, start with the first term and add the common difference each time:

$$t_1=11, \qquad t_n=t_{n-1}-3 \text{ for } n\ge2.$$

For an AP, if the 3rd term is 12 and the 50th term is 106, let first term be $a$ and common difference be $d$.

Then

$$a+2d=12$$
$$a+49d=106$$

Subtract:

$$47d=94 \Rightarrow d=2$$

Then

$$a+2(2)=12 \Rightarrow a=8$$

Now the 29th term is

$$t_{29}=a+28d=8+28\cdot 2=8+56=64$$

So the 29th term is 64.

The two-digit multiples of 3 form an AP:

$$12,15,18,\dots,99$$

Here $a=12$, $d=3$, and $l=99$.

Number of terms:

$$n=\frac{l-a}{d}+1=\frac{99-12}{3}+1=29+1=30.$$

So there are 30 such numbers.

Sum:

$$S_n=\frac{n}{2}(a+l)=\frac{30}{2}(12+99)=15\cdot 111=1665.$$

So the sum is 1665.

Harish's salary forms an AP:

• first salary $a=5,00,000$
• yearly increment $d=20,000$

After $n$ years, salary is

$$a+(n-1)d$$

Set it equal to ₹7,00,000:

$$5,00,000+(n-1)20,000=7,00,000$$
$$20,000(n-1)=2,00,000$$
$$n-1=10$$
$$n=11$$

So ₹7,00,000 is reached in the 11th year of salary, which means after 10 years of increments. The textbook wording expects the answer 10 years.

The marbles are arranged in 25 rows with 1, 2, 3, ..., 25 marbles.

So total marbles =

$$1+2+3+\cdots+25$$

Use

$$S_n=\frac{n(n+1)}{2}$$

with $n=25$:

$$S_{25}=\frac{25\cdot 26}{2}=25\cdot 13=325.$$

So the child uses 325 marbles.

The height after each bounce is multiplied by $\frac{3}{4}=0.75$.

Starting from 24 feet:

• 1st bounce: $24\times 0.75=18$
• 2nd bounce: $18\times 0.75=13.5$
• 3rd bounce: $13.5\times 0.75=10.125$
• 4th bounce: $10.125\times 0.75=7.59375$
• 5th bounce: $7.59375\times 0.75=5.6953125\approx 5.695$

So the sequence of heights is 18, 13.5, 10.125, 7.59375, 5.695....