NCERT Solutions for Class 9 Maths Chapter 8 - Predicting What Comes Next: Exploring Sequences and Progressions

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NCERT Solutions for Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions: Exercise Questions

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Think and Reflect (Page 174)

Question: Can you describe the pattern in each of the above sequences? Can you

predict the next few numbers in these sequences?

$\textbf{Answer:}$

(i)
Natural numbers sequence: 1, 2, 3, 4, 5, 6,....
Here, the rule is $a_n=n$, where $n$ is a natural number
So, the next terms are 7, 8, 9, 10, ...

(ii)
Odd numbers sequence: 1, 3, 5, 7, 9, 11, ....
Here, the rule is $a_n=2n-1$, where $n$ is a natural number
So, the next terms are 13, 15, 17, 19,.....

(iii)

Triangular number sequence: 1, 3, 6, 10, 15, 21

Here, the rule is $a_n=\frac{n(n+1)}2$

So, the next terms are 28, 36, 45, 55,....

(iv)

Square number sequence: 1, 4, 9, 16, 25, 36,...

Here, the rule is $a_n=n^2$, where $n$ is a natural number

So, the next terms are 49, 64, 81, 100, ...

Think and Reflect (Page 176)

Question: Can you think of any other kinds of sequences? List out five different types

of sequences and discuss their properties with your friends.

$\textbf{Answer:}$

Question 1. Find the first five terms of the sequence in which the $n^{\text {th }}$ term is given by
(i) $t_n=3 n-4$,
(ii) $t_n=2-5 n$, and
(iii) $t_n=n^2-2 n+3$ for $n \geq 1$.

$\textbf{Answer:}$

(i)
Given: $t_n=3 n-4$
When $n=1$, $t_1=3\times1-4=-1$
When $n=2$, $t_2=3\times2-4=2$
When $n=3$, $t_3=3\times3-4=5$
When $n=4$, $t_4=3\times4-4=8$
When $n=5$, $t_5=3\times5-4=11$

(ii)
Given: $t_n=2-5 n$
When $n=1$, $t_1=2-5\times1=-3$
When $n=2$, $t_2=2-5\times2=-8$
When $n=3$, $t_3=2-5\times3=-13$
When $n=4$, $t_4=2-5\times4=-18$
When $n=5$, $t_5=2-5\times5=-23$

(iii)
Given: $t_n=n^2-2 n+3$
When $n=1$, $t_1=1^2-2\times1+3=2$
When $n=2$, $t_2=2^2-2\times2+3=3$
When $n=3$, $t_3=3^2-2\times3+3=6$
When $n=4$, $t_4=4^2-2\times4+3=11$
When $n=5$, $t_5=5^2-2\times5+3=18$

Question 2. Find the $10^{\text {th }}$ and $15^{\text {th }}$ terms of the sequence $t_n=5 n-3$ for $n \geq 1$.

$\textbf{Answer:}$
Given: $t_n=5 n-3$ for $n \geq 1$
When $t=10,$ $t_{10}=5\times10-3=47$
When $t=15,$ $t_{15}=5\times15-3=72$

Question 3. Determine whether 97 and 172 are terms of the sequence $t_n=5 n-3$ for $n \geq 1$.

$\textbf{Answer:}$
Given:$t_n=5 n-3$ for $n \geq 1$
(i)
According to the question,
$5n-3=97$
$⇒5n=100$
$\therefore n=20$
So, 97 is the 20th term of the sequence.

(ii)
According to the question,
$5n-3=172$
$⇒5n=175$
$\therefore n=35$
So, 172 is the 35th term of the sequence.

Question 4. Which term of the sequence $t_n=5 n-3$ for $n \geq 1$ is 607?

$\textbf{Answer:}$

Given: $t_n=5 n-3$ for $n \geq 1$
According to the question,
$5n-3=607$
$⇒5n=610$
$\therefore n=122$
So, 607 is the 122nd term of the sequence.

Question 5. A sequence is given by the recursive rule $t_1=-5, t_{n+1}=t_n+3$ for $n \geq 1$.
Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

$\textbf{Answer:}$
Given: $t_1=-5, t_{n+1}=t_n+3$ for $n \geq 1$
When $n=1$, $t_2=t_1+3=-5+3=-2$
When $n=2$, $t_3=t_2+3=-2+3=+1$
When $n=3$, $t_4=t_3+3=1+3=4$
When $n=4$, $t_5=t_4+3=4+3=7$
So, the first 5 terms of the sequence are: $-5, -2, 1, 4, 7$
Here, the common difference, $d$, between terms is +3.
Also, first term, $a$= $-5$
We know that $t_n=a+(n-1)d$
According to the question,
$52=-5+(n-1)3$
$⇒3n-3=57$
$⇒3n=60$
$\therefore n=20$
So, 52 is a term of this sequence, and it is the 20th term.

Question 6. Let $\mathrm{T}_1=1, \mathrm{~T}_2=2, \mathrm{~T}_3=4$, and $\mathrm{T}_{\mathrm{n}}=\mathrm{T}_{\mathrm{n}-1}+\mathrm{T}_{\mathrm{n}-2}+\mathrm{T}_{\mathrm{n}-3}$ for $n \geq 4$. Find $\mathrm{T}_4, \mathrm{~T}_5, \mathrm{~T}_6, \mathrm{~T}_7$, and $\mathrm{T}_8$.

$\textbf{Answer:}$
Given: $\mathrm{T}_1=1, \mathrm{~T}_2=2, \mathrm{~T}_3=4$
$\mathrm{T}_{\mathrm{n}}=\mathrm{T}_{\mathrm{n}-1}+\mathrm{T}_{\mathrm{n}-2}+\mathrm{T}_{\mathrm{n}-3}$
So, $\mathrm{T}_4=\mathrm{T}_3+\mathrm{T}_2+\mathrm{T}_1=1+2+4=7$
$\mathrm{T}_5=\mathrm{T}_4+\mathrm{T}_3+\mathrm{T}_2=7+4+2=13$
$\mathrm{T}_6=\mathrm{T}_5+\mathrm{T}_4+\mathrm{T}_3=13+7+4=24$
$\mathrm{T}_7=\mathrm{T}_6+\mathrm{T}_5+\mathrm{T}_4=24+13+7=44$
$\mathrm{T}_8=\mathrm{T}_7+\mathrm{T}_6+\mathrm{T}_5=44+24+13=81$

Question 1. Find the $10^{\text {th }}$ and $26^{\text {th }}$ terms of the AP: $3,8,13,18, \ldots$.

$\textbf{Answer:}$
AP Sequence: $3,8,13,18, \ldots$
We know that the $n$th term of the AP is $a_n=a+(n-1) d$, where $a$ is the first term and $d$ is the common difference
Here, first term, $a$ = 3 and common difference, $d$ = 8 - 3 =5
So, $a_{10}=3+(10-1)5=3+45=48$
and $a_{26}=3+(26-1)5=3+125=128$

Question 2. Which term of the AP: $21,18,15, \ldots$ is –81?
Also, is 0 a term of this AP? Give reasons for your answer.

$\textbf{Answer:}$
AP Sequence: $21,18,15, \ldots$
We know that the $n$th term of the AP is $a_n=a+(n-1) d$, where $a$ is the first term and $d$ is the common difference
Here, first term, $a$ = 21 and common difference, $d=18-21=-3$
According to the question,
$21+(n-1)\times-3=-81$
$⇒-3n+3=-102$
$⇒-3n=-105$
$\therefore n=35$
So, –81 is the 35th term of the sequence.

Now,
According to the question,
$21+(n-1)\times-3=0$
$⇒-3n+3=-21$
$⇒-3n=-24$
$\therefore n=8$
So, 0 is the 8th term of the sequence.
Since $n$ is a positive integer, 0 is a term of the AP.

Question 3. Find the $n^{\text {th }}$ term of the AP: $11,8,5,2 \ldots$ Write the recursive rule for this AP.

$\textbf{Answer:}$
Given AP sequence: $11,8,5,2 \ldots$
First term, $a$ = 11
Common difference, $d=8-11=-3$
So, $a_n=11+(n-1)\times-3=11-3n+3=14-3n$

A recursive rule defines each term based on the preceding term.
For this AP, each term is obtained by adding the common difference $-3$ to the previous term. ⇒ $a_1 = 11$
⇒ $a_n = a_{n-1} - 3$, For $n \geq 2$

Question 4. An AP consists of 50 terms in which the $3^{\text {rd }}$ term is 12 and the last term is 106. Find the $29^{\text {th }}$ term.
(Hint: If ' $a$ ' is the first term and 'd' the common difference, then we arrive at the equations $a+2 d=12$ and $a+49 d=106$. Solve this pair of linear equations for ' $a$ ' and ' $d$ '.)

$\textbf{Answer:}$

Let the first term of the AP be $a$, and the common difference be $d$.
The total number of terms is $n=50$
So, $a_{50}=106$
The $3^{\text {rd }}$ term is $a_3=12$.
Using the $n^{\text {th }}$ term formula $a_n=a+(n-1) d$ :
$
\begin{aligned}
& \Rightarrow a+2 d=12 ----------(1) \\
& \Rightarrow a+49 d=106---------(2)
\end{aligned}
$
Subtracting Equation 1 from Equation 2, we get:
$
\begin{aligned}
& \Rightarrow(a+49 d)-(a+2 d)=106-12 \\
& \Rightarrow 47 d=94 \\
& \Rightarrow d=2
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow a+2(2)=12 \\
& \Rightarrow a+4=12 \\
& \Rightarrow a=8
\end{aligned}
$
Now, finding the $29^{\text {th }}$ term ( $a_{29}$ ):

$
\begin{aligned}
& \Rightarrow a_{29}=a+28 d \\
& \Rightarrow a_{29}=8+28(2) \\
& \Rightarrow a_{29}=8+56=64
\end{aligned}
$

Question 5. How many 2 -digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

$\textbf{Answer:}$

The 2-digit multiples of 3 form the AP: 12, 15, 18,......,99
Here, first term, $a= 12$ and common difference, $d=15-12=3$, and $l=99$
We know that the $n$th term of the AP is $a_n=a+(n-1) d$, where $a$ is the first term and $d$ is the common difference
According to the question,
$99=12+(n-1)3$
$3n-3=87$
$3n=90$
$\therefore n=30$
So, 30 2-digit numbers are divisible by 3.

We also know, sum of the numbers, $S_n=\frac{n}{2}(a+l)$, where $l$ is the last term
$⇒S_{30}=\frac{30}2(12+99)$
$⇒S_{30}=15\times111=1665$
So, their sum is 1665.

Question 6. Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

$\textbf{Answer:}$

Harish's salary in the AP structure: 500000, 520000, 540000, ......,700000
So, first term, $a=500000,$ last term, $l=700000,$ and common difference, $d=20000$
We know that the $n$th term of the AP is $a_n=a+(n-1) d$
According to the question,
$700000=500000+(n-1)20000$
$⇒(n-1)20000=200000$
$⇒n-1=10$
$\therefore n =11$
Since the first term is the starting salary, it is reached after (11 - 1) = 10 years.

Hence, after 10 years, his income reached ₹7,00,000.

Question 7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

$\textbf{Answer:}$

Total marbles the child uses are: 1 + 2 + 3 + ..... + 25
Here, first term, $a=1,$ last term, $l=25$ and common difference, $d=1$
So, sum of the numbers, $S_n=\frac{n}{2}(a+l)$
$⇒S_{25}=\frac{25}2(1+25)=\frac{25}2\times26=25\times13=325$

Hence, 325 marbles were used by the child in total.

Question 1. Find the $12^{\text {th }}$ term of a GP with common ratio 2, whose $8^{\text {th }}$ term is 192.

$\textbf{Answer:}$

Let the first term of the Geometric Progression (GP) be $a$, and the common ratio be $r=2$.
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}$
According to the question,
$192=a\times2^{8-1}$
$192=a\times2^7$
$192=a\times128$
$\therefore a= \frac{192}{128}=\frac32$
Now,
$a_{12}=\frac32\times2^{12-1}=\frac32\times2^{11}=3\times2^{10}=3\times1024=3072$

So, the 12th term of the GP is 3072.

Question 2. Find the $10^{\text {th }}$ and $n^{\text {th }}$ terms of the GP: $5,25,125, \ldots$.

$\textbf{Answer:}$

GP Sequence is: $5,25,125, \ldots$
Let the first term of the Geometric Progression (GP) be $a=5$,
and the common ratio be $r=\frac{25}5=5$
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}=5\times5^{n-1}=5^n$

Also, $a_{10}=5^{10}=9765625$

*Question 3. A sequence is given by the recursive rule $t_1=2, t_{n+1}=3 t_n-2$ for $n \geq 1$. Which term of the sequence is 730?

$\textbf{Answer:}$
Given: $t_1=2$ and $t_{n+1}=3 t_n-2$ for $n \geq 1$
$t_2=3(2)-2=4 $
$t_3=3(4)-2=10 $
$ t_4=3(10)-2=28$
$ t_5=3(28)-2=82 $
$ t_6=3(82)-2=244 $
$ t_7=3(244)-2=730$

Hence, the correct answer is that 730 is the $7^{\text{th}}$ term of the sequence.

Question 4. Which term of the GP: $2,6,18, \ldots$ is 4374? Write the explicit formula as well as the recursive formula for the $n^{\text {th }}$ term.

$\textbf{Answer:}$

Let the first term of the Geometric Progression (GP) be $a=2$, and the common ratio be $r=\frac62=3$
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}=2 \cdot 3^{n-1}$
$\begin{aligned} & \Rightarrow 2 \cdot 3^{n-1}=4374 \\ & \Rightarrow 3^{n-1}=2187\end{aligned}$
$\Rightarrow 3^{n-1}=3^7$
$\Rightarrow n-1=7$
$\therefore n =8$
Hence, 4374 is the $8^{\text{th}}$ term of the sequence.

The recursive formula defines each term relative to its predecessor:
$
\begin{aligned}
& \Rightarrow a_1=2 \\
& \Rightarrow a_n=3 a_{n-1} , \text { for } n \geq 2
\end{aligned}
$

Question 5. A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to $60 \%$ of the height from which it fell. It continues bouncing in this way - each time rising to $60 \%$ of the previous height.
(i) What height does the ball reach after the $5^{\text {th }}$ bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6{ }^{\text {th }}$ time?

$\textbf{Answer:}$

A ball is dropped from a height of 80 metres.
After 1st bounce, 60% of its height $=80\times\frac{60}{100}=48$ metres
After 2nd bounce, 60% of previous height, i.e., 48 metres $=48 \times\frac{60}{100}=28.8$

(i)
So, after the 5th bounce, the height it reaches
= Initial height × $\text{(0.6)}^5$
= $80\times(0.6)^5$
= $80 \times 0.07776$
= $6.2208$

(ii)

Initial drop: The ball falls from $80\text{ m}$ down to the ground ($1^{\text{st}}$ impact).
$1^{\text{st}}$ bounce: It travels up $h_1$ and falls down $h_1$ ($2^{\text{nd}}$ impact).
$2^{\text{nd}}$ bounce: It travels up $h_2$ and falls down $h_2$ ($3^{\text{th}}$ impact).
$3^{\text{rd}}$ bounce: It travels up $h_3$ and falls down $h_3$ ($4^{\text{th}}$ impact).
$4^{\text{th}}$ bounce: It travels up $h_4$ and falls down $h_4$ ($5^{\text{th}}$ impact).
$5^{\text{th}}$ bounce: It travels up $h_5$ and falls down $h_5$ ($6^{\text{th}}$ impact).
$\text{Total Distance} = H + 2h_1 + 2h_2 + 2h_3 + 2h_4 + 2h_5$
Using the sum formula for a finite geometric series $S_n = a \left(\frac{1 - r^n}{1 - r}\right)$ for the bouncing terms, where the first term is $2h_1 = 2(80)(0.6) = 96$ and the number of terms is $5$:
$\text{Total Distance} = 80 + 96 \left( \frac{1 - (0.6)^5}{1 - 0.6} \right)$
$\begin{aligned} & \Rightarrow \text { Total Distance }=80+96\left(\frac{1-0.07776}{0.4}\right) \\ & \Rightarrow \text { Total Distance }=80+\frac{96}{0.4} \times 0.92224 \\ & \Rightarrow \text { Total Distance }=80+240 \times 0.92224 \\ & \Rightarrow \text { Total Distance }=80+221.3376 \\ & \Rightarrow \text { Total Distance }=301.3376 \mathrm{~m}\end{aligned}$

Hence, the total vertical distance travelled by the ball by the time it hits the ground for the $6^{\text{th}}$ time is $301.3376\text{ m}$.

Question 6. Which term of the sequence $2,2 \sqrt{2}, 4, \ldots$ is 128?

$\textbf{Answer:}$

Let the first term of the Geometric Progression (GP) be $a=2$, and the common ratio be $r=\frac{2\sqrt2}2=\sqrt2$
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}$
According to the question,
$128=2\times(\sqrt2)^{n-1}$
$⇒(\sqrt2)^{n-1}=64$
$⇒(2^{\frac12})^{n-1}=2^6$
$⇒(2)^{\frac{n-1}2}=2^6$
$⇒\frac{n-1}2=6$
$⇒n-1=12$
$⇒\therefore n=13$

Hence, 128 is the $13^{\text{th}}$ term of the sequence.

Question 7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.

Look at Fig. 8.12 and try to answer the following questions.

(i) How many red squares are there in Stages 0 to 3?

(ii) Can you predict the number of red squares in Stages 4 and 5?

(iii) Can you find a rule for the number of red squares at the $n^{\text {th }}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.

(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3?
What will be the area of the red region in Stages 4 and 5?
Find the explicit as well as the recursive formula for the area of the red region at the $n^{\text {th }}$ stage.
What happens to this area as $n$, the number of stages, goes on increasing?

$\textbf{Answer:}$

(i)
At stage 0, there is 1 red square.
At stage 1, there are 8 red squares.
At stage 2, there are (8 × 8)= 64 red squares.
At stage 3, there are (64 × 8)= 512 red squares.

(ii)
$\text{Number of red squares in Stage 4} = 512 \times 8 = 4096$
$\text{Number of red squares in Stage 5} = 4096 \times 8 = 32768$

(iii)
Number of red squares at the $n^{\text{th}}$ stage, where $n\geq 0$ is: $8^n$
The recursive formula expressing each stage relative to its predecessor is:
$N_0 = 1$
⇒ $N_n = 8 \times N_{n-1}$ for $n \geq 1$

(iv)
Given:
The area of the square in Stage 0 is 1 square unit.
At each stage, only 8 out of the 9 smaller squares are retained.
Therefore, the area is multiplied by $\frac89$ at every stage.

So, area of stage 1 = $(\frac89)^1=\frac89$
Area of stage 2 = $(\frac89)^2=\frac{64}{81}$
Area of stage 3 = $(\frac89)^3=\frac{512}{729}$
Area of stage 4 = $(\frac89)^4=\frac{4096}{6561}$
Area of stage 5 = $(\frac89)^5=\frac{32768}{59049}$
These areas are all in a Geometrical Progression (GP) sequence.
So, $A_n=\left(\frac{8}{9}\right)^n$, where $n$ is the number of stage

Recursive Formula for Area is:
$A_0=1 $
$ A_n=\frac{8}{9} A_{n-1}$ for all $n \geq 1$

Since $0<\frac{8}{9}<1$, repeated multiplication by $\frac89$​ makes the area smaller and smaller.
Thus, the area of the red region approaches 0 as $n$ increases.

Question 1. Find the $31^{\text {st }}$ term of an AP whose $11^{\text {th }}$ term is 38 and $16^{\text {th }}$ term is 73.

$\textbf{Answer:}$

Let the first term of the Arithmetic Progression (AP) be $a$, and the common difference be $d$.
The formula for the $n^{\text{th}}$ term ($a_n$) of an AP is given by: $a_n = a + (n - 1)d$
According to the question,
$a+(11-1)d=38$
$⇒a+10d=38----(1)$
Also, $a+(16-1)d=73$
$⇒a+15d=-----(2)$
Subtracting Equation 1 from Equation 2, we get,
⇒ $(a + 15d) - (a + 10d) = 73 - 38$
⇒ $5d=35$
$\therefore d=7$
Putting the value of $d$, in equation 1 we get,
$a+10\times7=38$
$⇒a=-32$
Now, the 31st term of the sequence
$=-32+(31-1)\times7=-32+210=178$

Question 2. Determine the AP whose third term is 16 and whose $7^{\text {th }}$ term exceeds the $5^{\text {th }}$ term by 12.

$\textbf{Answer:}$
Let the first term of the AP be $a$, and the common difference be $d$.
The formula for the $n^{\text{th}}$ term ($a_n$) of an AP is given by: $a_n = a + (n - 1)d$
Given: Third term, $a_3$ is 16.
Also, $a_7-a_5=12$
$⇒a+6d-(a+4d)=12$
$⇒2d=12$
$\therefore d=6$
Also, $a + 2d = 16$ [Given]
$⇒a+2\times6=16$
$\therefore a =4$
Using $a = 4$ and $d = 6$, the AP sequence is: $ 4, 10, 16, 22, \dots$

*Question 3. How many three-digit numbers are divisible by 7?
(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)

$\textbf{Answer:}$

The three-digit numbers range from 100 to 999.
The smallest three-digit number divisible by 7 is $\frac{105}{7}=15$
The largest three-digit number divisible by 7 is $\frac{994}{7}=142$
AP sequence: $105,112,119, \ldots, 994$
where the first term is $a=105$, the common difference is $d=7$, and the last term is $a_n=994$.
The formula for the $n^{\text{th}}$ term ($a_n$) of an AP is given by: $a_n = a + (n - 1)d$
$
\begin{aligned}
& \Rightarrow 994=105+(n-1) 7 \\
& \Rightarrow 889=(n-1) 7 \\
& \Rightarrow n-1=\frac{889}{7}=127 \\
& \Rightarrow n=128
\end{aligned}
$
Hence, there are 128 three-digit numbers divisible by 7.

*Question 4. How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250 .)

$\textbf{Answer:}$
The smallest multiple of 4 greater than 10 is 12.
The largest multiple of 4 less than 250 is 248.
AP sequence: $12,16,20, \ldots, 248$
where the first term is $a=12$, the common difference is $d=4$, and the last term is $a_n=248$
The formula for the $n^{\text{th}}$ term ($a_n$) of an AP is given by: $a_n = a + (n - 1)d$
$
\begin{aligned}
& \Rightarrow 248=12+(n-1) 4 \\
& \Rightarrow 236=(n-1) 4 \\
& \Rightarrow n-1=\frac{236}{4}=59 \\
& \Rightarrow n=60
\end{aligned}
$

Hence, there are 60 multiples of 4 lying between 10 and 250.

*Question 5. Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

$\textbf{Answer:}$

Let the first term of the Geometric Progression (GP) be $a$, and the common ratio be $r$.
So, $a+ar=-4$
$⇒a(1+r)=-4-----(1)$
Also, $a_5=4a_3$
$⇒ar^4=4\times ar^2$
$⇒r^2=4$
$\therefore r=2,-2$
When $r=2$
Putting this in equation 1, we get,
$a(1+2)=-4$
$⇒a=-\frac43$
GP Sequence is: $-\frac{4}{3},-\frac{8}{3},-\frac{16}{3}, \ldots$
When $r=-2$
Putting this in equation 1, we get,
$a(1-2)=-4$
$⇒-a=-4$
$\therefore a=4$
GP Sequence is: $4, -8, 16, -32, \dots$

*Question 6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

$\textbf{Answer:}$

Let 100 be expressed as a sum of $k$ consecutive natural numbers starting from $a$.
So, $a+(a+1)+(a+2)+\cdots+(a+k-1)=100$
Using the sum formula of an arithmetic sequence, we get,
$⇒\frac{k}{2}(2 a+k-1)=100$
$⇒k(2a+k-1)=200$
$⇒2a+k-1=\frac{200}k$
$\therefore a=\frac{\frac{200}{k}-k+1}{2}$, where $a$ and $k$ are positive integers

If $k=5,$ $a=\frac{40-5+1}{2}=18$
So, $100=18+19+20+21+22$

If $k=8,$ $a=\frac{25-8+1}{2}=9$
So, $100=9+10+11+12+13+14+15+16$
No other divisors give a positive integer $a$.

*Question 7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour?

$\textbf{Answer:}$

Given: Initial number of bacteria, $a$ = 30
Since the population doubles every hour, the common ratio is $r = 2$.
The number of bacteria after $n$ hours, $T_n=30\left(2^n\right)$
At the 2nd hour, total bacteria $=30\times2^2=120$
At the 4th hour, total bacteria $=30\times2^4=480$
At the $n$th hour, total bacteria $=30\times2^n$

*Question 8. The sum of the $4^{\text {th }}$ and $8^{\text {th }}$ terms of an AP is 24 and the sum of the $6^{\text {th }}$ and $10^{\text {th }}$ terms is 44. Find the first three terms of the AP.

$\textbf{Answer:}$

Let the first term of the Arithmetic Progression (AP) be $a$ and the common difference be $d$.
The general formula for the $n^{\text{th}}$ term is $a_n = a + (n-1)d$
According to the question,
$a_4+a_8=24$
$⇒a+3d+a+7d=24$
$⇒2a+10d=24$
$⇒a+5d=12------(1)$
Also,
$a_6+a_{10}=44$
$⇒a+5d+a+9d=44$
$⇒2a+14d=44$
$⇒a+7d=22-----(2)$
Subtracting Equation 1 from Equation 2, we get,
$(a + 7d) - (a + 5d) = 22 - 12$
$⇒2d=10$
$\therefore d=5$
Putting it in equation 1, we get,
$a+5\times5=12$
$⇒a=-13$
So, 2nd term = $-13+5=-8$
Third term = $-8+5=-3$

Hence, the first three terms of the AP are $-13, -8, \text{ and } -3$.

*Question 9. Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than 1,000.

$\textbf{Answer:}$

Sum of the first $n$ natural numbers, $S_n = \frac{n(n + 1)}{2}$
We need to find the smallest integer value of $n$ such that $S_n > 1000$.
$⇒\frac{n(n + 1)}{2}>1000$
$⇒n(n+1)>2000$
Let's find the approximate root by checking perfect squares near 2000.
We know $45^2 = 2025$
For $n=44: 44 \times 45=1980$ (which is less than 2000)
For $n=45: 45 \times 46=2070$ (which is greater than 2000)
Now,
$S_{44}=\frac{44 \times 45}{2}=990$
$ S_{45}=\frac{45 \times 46}{2}=1035$, which is greater than 1000

Hence, the smallest value of $n$ is 45.

*Question 10. Which term of the GP: $2,8,32, \ldots$ is 131072?
Write the explicit formula as well as the recursive formula for the $n^{\text {th }}$ term.

$\textbf{Answer:}$

GP Sequence: $2,8,32, \ldots$
Here, first term, $a = 2$ and common ratio,$r = \frac{8}{2} = 4$.
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}$
$⇒a_n = 2 \times4^{n-1}$

The recursive formula defines each term relative to its predecessor:
⇒ $a_1 = 2$
⇒ $a_n = 4a_{n-1}$ for $n \ge 2$

According to the question,
$131072=2 \times4^{n-1}$
$⇒4^{n-1}=65536$
$⇒4^{n-1}=4^8$
$⇒n-1=8$
$\therefore n=9$
Hence, 131072 is the $9^{\text{th}}$ term.

*Question 11. The sum of the first three terms of a GP is $\frac{13}{12}$ and their product is -1. Find the common ratio and the terms.

$\textbf{Answer:}$

Let the first three terms of the GP be written in the form $\frac{a}{r}, a, \text{ and } ar$,
where $a$ is the middle term and $r$ is the common ratio.
According to the question,
$\frac ar\times a \times ar=-1$
$a^3=-1$
$\therefore a=-1$
Also,
$\frac{a}{r}+a+ar=\frac{13}{12}$
$⇒-\frac{1}{r}-1-r=\frac{13}{12}$
$⇒-1-r-r^2=\frac{13 r}{12}$
$⇒12 r^2+25 r+12=0 $
$⇒ (3 r+4)(4 r+3)=0$
$⇒r=-\frac43,-\frac34$
When $r=-\frac43$,
Terms of the GP are: $\frac{-1}{-\frac{4}{3}},-1, \frac{4}{3}⇒\frac34,-1,\frac43$

When $r=-\frac34$,
Terms of the GP are: $\frac{4}{3},-1, \frac{3}{4}$

*Question 12. If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a GP are $x, y$ and $z$ respectively, prove that $x, y, z$ are in GP.

$\textbf{Answer:}$

Let the first term of the GP be $a$, and the common ratio be $r$.
The formula for the $n^{\text {th }}$ term of a GP is given by: $a_n=a r^{n-1}$
So, $a_4=ar^3=x$
Also, $a_{10}=ar^9=y$ and $a_{16}=a^{15}=z$
For three numbers $x, y, z$ to be in a Geometric Progression, the ratio of consecutive terms must be equal. ($y^2 = xz$)
Here, $y^2=a^2(r^9)^2=a^2r^{18}$
Also, $xz=ar^3\times a^{15}=a^2r^{18}$
Thus, $y^2=xz$

Hence, $x, y, z$ are in GP.

*Question 13. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

$\textbf{Answer:}$

Let the first three terms of the GP be written in the form $a, ar,\text{ and } ar^2$,
where $a$ is the middle term and $r$ is the common ratio.
According to the question,
$a + ar + ar^2 = 26$
$⇒a(1+r+r^2)=26$
Squaring both sides, we get,
$⇒a^2(1 + r + r^2)^2 = 26^2 = 676 -----------(1)$
Also,
$a^2 + a^2r^2 + a^2r^4 = 364$
$⇒a^2(1+r^2+r^4)=364-----------(2)$
Dividing Equation 1 by Equation 2, we get,
$\frac{a^2\left(1+r+r^2\right)^2}{a^2\left(1+r^2+r^4\right)}=\frac{676}{364}$
$⇒\frac{\left(1+r+r^2\right)^2}{\left(1+r+r^2\right)\left(1-r+r^2\right)}=\frac{13}{7}$
$⇒\frac{1+r+r^2}{1-r+r^2}=\frac{13}{7}$
$⇒7\left(1+r+r^2\right)=13\left(1-r+r^2\right)$
$ ⇒ 7+7 r+7 r^2=13-13 r+13 r^2$
$⇒6r^2-20r+6=0$
$⇒3r^2-10r+3=0$
$⇒ (3r - 1)(r - 3) = 0$
$⇒ r = 3 \text{ or } r = \frac{1}{3}$
When $r=3$,
Putting the value in equation 1, we get,
$a^2(1+3+9)^2=676$
$⇒a^2\times 169=676$
$a^2=4$
$a=2$
So, terms in the GP are: 2, 6, 18, 54, ....

When $r=\frac13$,
Putting the value in equation 1, we get,
$a^2\left(1+\frac{1}{3}+\frac{1}{9}\right)^2=676$
$⇒a^2\times \frac{169}{81}=676$
$a^2=324$
$a=18$
The terms are: $18, 18\left(\frac{1}{3}\right), 18\left(\frac{1}{3}\right)^2,... \Rightarrow 18, 6, 2,...$

*Question 14. Suppose $\mathrm{P}_1=1, \mathrm{P}_2=2$ and for $n>2, \mathrm{P}_{\mathrm{n}}=\mathrm{P}_1+\mathrm{P}_2+\cdots+\mathrm{P}_{\mathrm{n}-1}+1$. Find the values of $\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_8$.
Can you find a simpler recursive formula for $\mathrm{P}_{\mathrm{n}}$? Can you give an explicit formula?

$\textbf{Answer:}$

Given: $\mathrm{P}_1=1, \mathrm{P}_2=2$, and $\mathrm{P}_{\mathrm{n}}=\mathrm{P}_1+\mathrm{P}_2+\cdots+\mathrm{P}_{\mathrm{n}-1}+1$
$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 $
$P_5=P_1+P_2+P_3+P_4+1=1+2+4+8+1=16$
$P_6=1+2+4+8+16+1=32 $
$P_7=1+2+4+8+16+32+1=64$
$P_8=1+2+4+8+16+32+64+1=128$

Here, each term is twice the previous term.
So, the simpler recursive formula is:
$P_1=1$
$P_n=2 P_{n-1}$ For all $n \geq 2$

The values computed are $P_1 = 1$, $P_2 = 2^1$, $P_3 = 2^2$, $P_4 = 2^3$, and so on.
So, the explicit formula for any index $n$ is $P_n = 2^{n-1}$ $\text{for all } n \ge 1$

*Question 15. Suppose $\mathrm{W}_1=1, \mathrm{~W}_2=2$ and for $n>2, \mathrm{~W}_{\mathrm{n}}=\mathrm{W}_1+\mathrm{W}_2+\cdots+ \mathrm{W}_{\mathrm{n}-2}+2$.
Find the values of $\mathrm{W}_1, \mathrm{~W}_2, \ldots, \mathrm{~W}_8$.
Do you recognise this sequence?

$\textbf{Answer:}$

We are given $W_1=1, W_2=2$, and the relation $W_n=W_1+W_2+\cdots+W_{n-2}+2$ for $n>2$.
Now,
$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=W_1+W_2+W_3+W_4+2=1+2+3+5+2=13$
$ W_7=W_1+W_2+W_3+W_4+W_5+2=1+2+3+5+8+2=21$
$ W_8=W_1+W_2+W_3+W_4+W_5+W_6+2=1+2+3+5+8+13+2=34$
So, the sequence is: 1, 2, 3, 5, 8, 13, 21, 34, ...
We can see,
3 = 1 + 2, 5 = 2 + 3, 8 = 3 + 5, and so on
So, $W_n=W_{n-1}+W_{n-2}$
This is the Fibonacci sequence[starting with 1, 2 instead of 1, 1].

Predicting What Comes Next: Exploring Sequences and Progressions Class 9 Chapter 8: Topics

Topics you will learn in NCERT Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions include:

• 8.1 Introduction to Sequences
• 8.2 Explicit Rule for a Sequence
• 8.3 Recursive Rule for a Sequence
• 8.4 Arithmetic Progressions
• 8.5 Sum of the First n Natural Numbers
• 8.6 Geometric Progressions

Class 9 Maths NCERT Chapter 8 Solutions: Extra Questions

Question 1:
Which of the following statements is correct?
I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$ is 4840.
II. The value of $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$ is $\mathrm{k}^{16} - \frac{1}{\mathrm{k}^{16}}$.

$\textbf{Answer:}$
I. $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$
$= (100-99)(100+99) + (98-97)(98+97) + .......... + (22-21)(22+21)$
$= (1)(199) + (1)(195) + ............. + (1)(43)$
Here, $a$ = 43 and $d$ = 4
Let the number of terms be $n$.
So, $a + (n-1)d = 199$
$⇒ 43 + 4(n-1) = 199$
$⇒ 4(n-1) = 156$
$⇒ n-1 = 39$
$⇒ n = 40$
Sum of AP = $\frac{n}{2}(2a + (n-1)d)=\frac{40}{2}(43 + 199)= 4840$
II. $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^2-\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
= $\left(\mathrm{k}^8-\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$
Hence, the correct answer is "only statement I is correct".

Question 2:

Find the sum of $3+3^2+3^3+\ldots+3^8$.

$\textbf{Answer:}$

The given sequence represents GP of constant ratio $r$ = 3,
The sum of a Geometric progression (GP) of $n$ terms with first term $a$ and common ratio equal to $r$ = $\frac{a\times{(r^{n}-1)}}{r-1}$
According to the question $n = 8, a = 3$
$\therefore$ Sum = $\frac{3\times{(3^{8}-1)}}{3-1}$ = 9840.
Hence, the correct answer is 9840.

Question 3:

A bus starts running with an initial speed of 21 km/hr, and its speed increases by 3 km/hr every hour. How many hours will it take to cover a distance of 252 kilometres?

$\textbf{Answer:}$

Given: Initial speed ($a$) = 21 km/hr
Increase in speed every hour ($d$) = 3 km/hr
This is an arithmetic progression, with the initial number ($a$) = 21 and the common difference ($d$) = 3
Also, the sum of A.P. = $\frac{n}{2}[2a + (n - 1)d]$
⇒ 252 = $\frac{n}{2}[2×21+(n-1)3]$
⇒ $3n^2+ 39n - 504 = 0$
⇒ $n^2+13n-168=0$
⇒ $(n – 8)(n + 21)=0$
⇒ $n = 8, –21$
Since the number of hours can’t be negative, the time taken is 8 hours.
Hence, the correct answer is 8.

Question 4:

What is the value of $14^{3}+16^{3}+18^{3}+...+30^{3}$?

$\textbf{Answer:}$

Given:
$14^{3}+16^{3}+18^{3}+.....+30^{3}$
$= 2^{3}\times7^{3}+2^{3}\times8^{3}+2^{3}\times9^{3}+.....+2^{3}\times15^{3}$
$= 2^{3}(7^{3}+8^{3}+9^{3}+.....+15^{3})$
$= 2^{3}[(1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3})-(1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3})]$
The sum of cubes of $n$ natural numbers$ =\left[\frac{n(n+1)}{2}\right]^2$
For $n = 6$, we have
$1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}=\left[\frac{6(6+1)}{2}\right]^2 = \left[\frac{42}{2}\right]^2 = (21)^{2} =$ 441
For $n = 15$, we have
$1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3}=\left[\frac{15(15+1)}{2}\right]^2 = \left[\frac{240}{2}\right]^2 = (120)^{2} =$ 14400
Now, $2^{3}[(1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3})-(1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3})]$
$=2^{3}(14400 - 441)$
$= 8 × 13959$
$= 111672$
Hence, the correct answer is 111672.

Question 5:

If $A =(\frac{1}{0.4})+(\frac{1}{0.04})+(\frac{1}{0.004})+....$ upto 8 terms, then what is the value of $A$?

$\textbf{Answer:}$

Given: $ A =(\frac{1}{0.4})+(\frac{1}{0.04})+(\frac{1}{0.004})+....$ upto 8 terms.
This expression can be written as:
$⇒ A =(\frac{10}{4})+(\frac{100}{4})+(\frac{1000}{4})+(\frac{10000}{4})+(\frac{100000}{4})+(\frac{1000000}{4})(\frac{10000000}{4})+(\frac{100000000}{4})$
Using the formula, $S_n = \frac{a(r^n-1)}{r-1}$, where $a$ = first term, $r$ = common ratio, $n$ = number of terms
$10 + 100 + 1000+ .............+ 100000000 = \frac{10(10^8-1)}{10-1}=111111110$
$⇒ A=(\frac{111111110}{4})$
$\therefore A = 27777777.5$
Hence, the correct answer is 27777777.5.

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Class 9 Books & Syllabus

Before the start of a new academic year, students should refer to the latest syllabus to determine the chapters they’ll be studying. Below are the updated syllabus links, along with some recommended reference books.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9