Exploring Sequences and Progressions

Exploring Sequences and Progressions – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook.

NCERT Solutions Class 9

Exploring Sequences and Progressions – NCERT Solutions

Q.1:

Why is it useful to have an explicit formula for the nth term of a sequence?

Solution:

It is useful to have an explicit formula for the nth term because:

• It helps us find any term directly without writing all the previous terms.
• It saves time and effort, especially for large values of n.
• It helps in identifying patterns clearly (like linear patterns such as 2n – 1).
• It allows us to solve problems easily, such as finding which term has a given value.

An explicit formula makes working with sequences faster, easier, and more systematic.

Q.2:

Can you find the rule describing the nth term of the sequence of square numbers?

Solution:

Yes.
The sequence of square numbers is:

$1,4,9,16,25, \ldots$
Observe:

• $1=1^2$
• $4=2^2$
• $9=3^2$
• $16=4^2$
• $25=5^2$

So, the rule is:
nth term $=n^2$
The $n$th term of the sequence of square numbers is $n^2$.

Q.3:

Find the first five terms of the sequence in which the $n^{\text {th }}$ term is given by $t_n=3 n-4$.

Solution:

$t_n=3 n-4$

• $n=1: 3(1)-4=-1$
• $n=2: 6-4=2$
• $n=3: 9-4=5$
• $n=4: 12-4=8$
• $n=5: 15-4=11$

First five terms:

$-1,2,5,8,11$

Q.4:

Find the first five terms of the sequence in which the $n^{\text {th }}$ term is given by $t_n=2-5 n$.

Solution:

$t_n=2-5 n$

• $n=1: 2-5=-3$
• $n=2: 2-10=-8$
• $n=3: 2-15=-13$
• $n=4: 2-20=-18$
• $n=5: 2-25=-23$

First five terms:

$-3,-8,-13,-18,-23$

Q.5:

Find the first five terms of the sequence in which the $n^{\text {th }}$ term is given by $t_n=n^2-2 n+3$.

Solution:

$t_n=n^2-2 n+3$

• $n=1: 1-2+3=2$
• $n=2: 4-4+3=3$
• $n=3: 9-6+3=6$
• $n=4: 16-8+3=11$
• $n=5: 25-10+3=18$

First five terms:

$2,3,6,11,18$

Q.6:

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

Solution:

Given: $t_n=5 n-3$
Find: 10th term

$t_{10}=5(10)-3=50-3=47$
15th term

$t_{15}=5(15)-3=75-3=72$
10th term = 47
15th term $=72$

Q.7:

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

Solution:

Given: $t_n=5 n-3$
To check if a number is a term, set $5 n-3=$ given number and solve for $n$.

For 97:

$5 n-3=97$
$5 n=100$
$n=20 \text { (a whole number) }$
So, 97 is a term (20th term).

For 172:

$5 n-3=172$
$5 n=175$
$n=35 \text { (a whole number) }$
So, 172 is also a term (35th term).

Q.8:

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

Solution:

Given: $t_n=5 n-3$
We need to find $n$ such that the term is 607.
Set:

$5 n-3=607$
$5 n=610$
$n=122$
607 is the 122nd term of the sequence.

Q.9:

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?

Solution:

Given:

$t_1=-5, t_{n+1}=t_n+3$
This means each term increases by 3.
First five terms:

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

So, first five terms:

$-5,-2,1,4,7$
Now check if 52 is a term:
General term:

$t_n=-5+(n-1) \times 3$
Set equal to 52:

$-5+3(n-1)=52$
$3(n-1)=57$
$n-1=19$
$n=20$
Yes, 52 is a term, and it is the 20th term of the sequence.

Q.10:

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

Solution:

Given:
$T_1=1, T_2=2, T_3=4$
and $T_n=T_{n-1}+T_{n-2}+T_{n-3}$
Now find terms step by step:

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

$T_4=7, T_5=13, T_6=24, T_7=44, T_8=81$

Q.11:

Find the 10^th and 26^th terms of the AP: 3, 8, 13, 18, …

Solution:

Given AP: $3,8,13,18, \ldots$
First term $a=3$, common difference $d=5$
Use formula:

$t_n=a+(n-1) d$
10th term:

$t_{10}=3+(10-1) \times 5=3+45=48$
26th term:

$t_{26}=3+(26-1) \times 5=3+125=128$
10th term = 48
26th term = 128

Q.12:

Which term of the AP : 21, 18, 15, … is -81? Also, is 0 a term of this AP? Give reasons for your answer.

Solution:

Given AP: $21,18,15, \ldots$
First term $a=21$, common difference $d=-3$
General term:

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

Find which term is -81:

$21+(n-1)(-3)=-81$
$21-3(n-1)=-81$
$21-3 n+3=-81$
$24-3 n=-81$
$-3 n=-105$
$n=35$
So, -81 is the 35th term.
Check if 0 is a term:

$21+(n-1)(-3)=0$
$21-3(n-1)=0$
$24-3 n=0$
$3 n=24$
$n=8$
Since $n$ is a whole number, 0 is a term.
-81 is the 35th term.
Yes, 0 is the 8th term of the AP.

Q.13:

Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Solution:

Given AP: $11,8,5,2, \ldots$
First term $a=11$, common difference $d=-3$
nth term:

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

Recursive rule:

$t_1=11, t_n=t_{n-1}-3 \text { for } n \geq 2$
Final Answer:

$t_n=14-3 n$
Recursive rule: $t_1=11, t_n=t_{n-1}-3$

Q.14:

An AP consists of 50 terms in which the 3^rd term is 12 and the last term is 106. Find the 29^th term.

Solution:

Given:

• $t_3=a+2 d=12$
• $t_{50}=a+49 d=106$

Subtract the first equation from the second:

$(a+49 d)-(a+2 d)=106-12$
$47 d=94 \Rightarrow d=2$
Now substitute $d=2$ in $a+2 d=12$:

$a+4=12 \Rightarrow a=8$
Now find the 29th term:

$t_{29}=a+(29-1) d=8+28 \times 2=8+56=64$
29th term = 64

Q.15:

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

Solution:

Two-digit numbers range from 10 to 99.
Numbers divisible by 3 in this range form an AP:

$12,15,18, \ldots, 99$
Here, $a=12, d=3$, last term $l=99$

Number of terms:

$n=\frac{l-a}{d}+1=\frac{99-12}{3}+1=29+1=30$
So, there are 30 numbers.

Sum of these numbers:

$S=\frac{n}{2}(a+l)=\frac{30}{2}(12+99)=15 \times 111=1665$
Number of terms = 30
Sum = 1665

Q.16:

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?

Solution:

Harish’s salary forms an AP:

• First term $a=5,00,000$
• Common difference $d=20,000$
• Required term $=7,00,000$

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

$7,00,000=5,00,000+(n-1) \times 20,000$
$7,00,000-5,00,000=(n-1) \times 20,000$
$2,00,000=(n-1) \times 20,000$
$n-1=10 \Rightarrow n=11$
This means salary becomes ₹ 7,00,000 in the 11th year.
Since he starts at year 1, the number of years taken:

$11-1=10$
His income reached ₹ 7,00,000 after 10 years.

Q.17:

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?

Solution:

Number of marbles in rows:

$1,2,3, \ldots, 25$
This is an AP with:
$a=1, d=1, n=25$

Sum of first $n$ terms:

$S_n=\frac{n}{2}(a+l)$
Here, last term $l=25$

$S_{25}=\frac{25}{2}(1+25)=\frac{25}{2} \times 26=25 \times 13=325$
Total marbles = 325

Q.18:

Find the 12^th term of a GP with common ratio 2, whose 8th term is 192.

Solution:

For a GP,

$t_n=a r^{n-1}$
Given:

• Common ratio $r=2$
• $t_8=192$

$t_8=a r^7=a \cdot 2^7=128 a=192$
$a=\frac{192}{128}=\frac{3}{2}$
Now find 12th term:

$t_{12}=a r^{11}=\frac{3}{2} \cdot 2^{11}=\frac{3}{2} \cdot 2048=3 \cdot 1024=3072$
12th term = 3072

Q.19:

Find the 10^th and n^th terms of the GP: 5, 25, 125, …

Solution:

Given GP: $5,25,125, \ldots$
First term $a=5$, common ratio $r=5$
nth term:

$t_n=a \cdot r^{n-1}=5 \cdot 5^{n-1}=5^n$
10th term:

$t_{10}=5^{10}=9765625$

Q.20:

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?

Solution:

Given:

$t_1=2, t_{n+1}=3 t_n-2$
Find terms until we reach 730:

• $t_1=2$
• $t_2=3(2)-2=4$
• $t_3=3(4)-2=10$
• $t_4=3(10)-2=28$
• $t_5=3(28)-2=82$
• $t_6=3(82)-2=244$
• $t_7=3(244)-2=730$

730 is the 7th term of the sequence.

Q.21:

Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.

Solution:

Given GP: $2,6,18, \ldots$
First term $a=2$, common ratio $r=3$

Explicit formula:

$t_n=a \cdot r^{n-1}=2 \cdot 3^{n-1}$
Find which term is 4374:

$2 \cdot 3^{n-1}=4374$
$3^{n-1}=2187=3^7$
$n-1=7 \Rightarrow n=8$
So, 4374 is the 8th term.

Recursive formula:

$t_1=2, t_n=3 t_{n-1}(n \geq 2)$
Final Answer:
4374 is the 8th term
Explicit: $t_n=2 \cdot 3^{n-1}$
Recursive: $t_1=2, t_n=3 t_{n-1}$

Q.22:

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.

1. What height does the ball reach after the 5th bounce?
2. What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

Solution:

Given:
Initial height $=80 {~m}$
Each bounce reaches $60 \%=0.6$ of previous height

1. Height after 5th bounce Heights form a GP:
   $80 \times(0.6)^n$
   After 5th bounce: $h_5=80 \times(0.6)^5=80 \times 0.07776=6.2208 {~m}$
   So, height $\approx 6.22 {~m}$
2. Total distance till it hits ground 6th time Motion:
   First fall: 80 m
   Then for each bounce: goes up and comes down Heights after each bounce: $h_1=48, h_2=28.8, h_3=17.28, h_4=10.368, h_5=6.2208$
   Total distance: $\text { Distance }=80+2\left(h_1+h_2+h_3+h_4+h_5\right)$
   $=80+2(48+28.8+17.28+10.368+6.2208)$
   $=80+2(110.6688)=80+221.3376=301.3376 {~m}$

Q.23:

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

Solution:

Given sequence:

$2,2 \sqrt{2}, 4, \ldots$
This is a GP with:

• First term $a=2$
• Common ratio $r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$

So,

$t_n=a \cdot r^{n-1}=2(\sqrt{2})^{n-1}$
We need:

$2(\sqrt{2})^{n-1}=128$
Write in powers of 2:

$\sqrt{2}=2^{1 / 2}$
$2\left(2^{1 / 2}\right)^{n-1}=128$
$2 \cdot 2^{(n-1) / 2}=128=2^7$
$2^{1+\frac{n-1}{2}}=2^7$
$1+\frac{n-1}{2}=7$
$\frac{n-1}{2}=6 \Rightarrow n-1=12 \Rightarrow n=13$

Q.24:

Figure 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 the figure and try to answer the following questions.

1. How many red squares are there in Stages 0 to 3? (1)
2. Can you predict the number of red squares in Stages 4 and 5? (1)
3. Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (2) OR 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 nth stage. What happens to this area as n, the number of stages, goes on increasing? (2)

Solution:

1. Number of red squares in Stages 0 to 3: Stage 0 = 1
   Stage $1=8$
   Stage $2=8 \times 8=64$
   Stage $3=8 \times 8 \times 8=512$
   So, the sequence is: $1,8,64,512$
2. Stage 4 and Stage 5: Stage $4=8^4=4096$
   Stage $5=8^5=32768$
3. Rule (explicit and recursive): This is a geometric progression with common ratio 8.
   Explicit formula:
   ${t}_{{n}}=8^{{n}}$ (if Stage 0 is ${n}=0$ )
   So:
   Stage $n \rightarrow t_n=8^n$
   Recursive formula:
   ${t}_0=1$
   ${t}_{{n}}=8 \times {t}_{{n}-1}$, for ${n} \geq 1$ OR
   Each stage keeps 8 out of 9 equal parts, so area multiplies by $8 / 9$ each time.
   • Area of red region: Stage 0 = 1
     Stage 1 = 8/9
     Stage $2=(8 / 9)^2=64 / 81$
     Stage $3=(8 / 9)^3=512 / 729$
   • Stage 4 and 5: Stage $4=(8 / 9)^4$
     Stage $5=(8 / 9)^5$
   • Formula: Explicit formula: $s_n=(8 / 9)^n$
     Recursive formula:
     ${s}_0=1$
     $s_n=(8 / 9) \times s_{n-1}$, for $n \geq 1$

Q.25:

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Solution:

We are given:
11th term:

\begin{equation*} a+10 d=38 \end{equation*} …(1)
16th term:

\begin{equation*} a+15 d=73 \end{equation*} …(2)
Subtract (1) from (2):

$(a+15 d)-(a+10 d)=73-38$
$5 d=35$
$d=7$
Now put ${d}=7$ in (1):

$a+10 \times 7=38$
$a+70=38$
$a=-32$
Now find 31st term:

$a_{31}=a+30 d$
$a_{31}=-32+30 \times 7$
$a_{31}=-32+210$
$a_{31}=178$

Q.26:

Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Solution:

We are given:

3rd term:
a + 2d = 16 …(1)

7th term exceeds 5th term by 12:

(a + 6d) – (a + 4d) = 12
2d = 12
d = 6

Now put d = 6 in equation (1):

a + 2 × 6 = 16
a + 12 = 16
a = 4

So, the AP is:

a = 4, d = 6

Hence the AP is:
4, 10, 16, 22, 28, …

Q.27:

How many three-digit numbers are divisible by 7?

Solution:

All 3-digit numbers divisible by 7 form an AP:
First 3-digit multiple of 7:
$100 \div 7=14$ remainder, so next multiple $=15 \times 7=105$
So, first term ${a}=105$
Largest 3-digit multiple of 7:
$999 \div 7=142$ remainder, so largest multiple $=142 \times 7=994$
So, last term l = 994
Common difference ${d}=7$
Now, number of terms:

$n=\frac{l-a}{d}+1$
$n=\frac{994-105}{7}+1$
$n=\frac{889}{7}+1$
$n=127+1$
$n=128$

Q.28:

How many multiples of 4 lie between 10 and 250?

Solution:

We need multiples of 4 between 10 and 250.
First multiple of 4 greater than 10:

$12=3 \times 4$
So, first term ${a}=12$
Last multiple of 4 less than 250:

$248=62 \times 4$
So, last term l $=248$
Common difference ${d}=4$
Now number of terms:

$n=(l-a) / d+1$
$n=(248-12) / 4+1$
$n=236 / 4+1$
$n=59+1$
$n=60$
Final answer: 60 multiples of 4 lie between 10 and 250.

Q.29:

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

Solution:

Let the GP be:
${a}, {ar}_{,} {ar}^2, {ar}^3, {ar}^4, \ldots$
Given:

Sum of first two terms $=-4$

a + ar = -4
a(1 + r) = -4 …(1)
Also, fifth term is 4 times the third term:

$a r^4=4\left(a r^2\right)$
Divide both sides by ${ar}^2({a} \neq 0, {r} \neq 0)$:

$r^2=4$
So, ${r}=2$ or ${r}=-2$

Now solve both cases:

Case 1: ${r}=2$
From (1):

$a(1+2)=-4$
$3 a=-4$
$a=-4 / 3$
So GP:

$-4 / 3,-8 / 3,-16 / 3,-32 / 3, \ldots$
Case 2: ${r}=-2$
From (1):

$a(1-2)=-4$
$-a=-4$
$a=4$
So GP:

$4,-8,16,-32, \ldots$

Q.30:

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

Solution:

Let the number 100 be expressed as sum of $n$ consecutive natural numbers starting from a:
${a}+({a}+1)+({a}+2)+\ldots+{n}$ terms

$\text { Sum }=\frac{n}{2}[2 a+(n-1)]=100$
So,

$n(2 a+n-1)=200$
$2 a n=200-n(n-1)$
Now we try values of $n$ such that $a$ is a natural number.

$n=1:$
$a=100$
$\text { So, } 100$
$n=5$
$5 a+10=100$
$5 a=90 \rightarrow a=18$
$\text { So, } 18+19+20+21+22$
$n=8:$
$8 a+28=100$
$8 a=72 \rightarrow a=9$
$\text { So, } 9+10+11+12+13+14+15+16$
For other values of ${n}, {a}$ is not a natural number.

Q.31:

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 2nd hour, 4th hour and nth hour?

Solution:

The number of bacteria doubles every hour, so it forms a GP:
Initial bacteria = 30
So,

$a=30$
$r=2$
General term:

$a_n=a \cdot r^{n-1}$
End of 2nd hour:

$a_2=30 \cdot 2^{2-1}=30 \cdot 2=60$
End of 4th hour:

$a_4=30 \cdot 2^{4-1}=30 \cdot 2^3=30 \cdot 8=240$
End of nth hour:

$a_n=30 \cdot 2^{n-1}$

• 2nd hour = 60 bacteria
• 4th hour $=240$ bacteria
• nth hour $=30 \cdot 2^{n-1}$ bacteria

Q.32:

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

Let the AP be:
a, a + d, a + 2d, …

Given:

4th term = a + 3d
8th term = a + 7d

So,
(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 …(1)

Also:

6th term = a + 5d
10th term = a + 9d

So,
(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 …(2)

Now subtract (1) from (2):

(a + 7d) – (a + 5d) = 22 – 12
2d = 10
d = 5

Now put d = 5 in (1):

a + 5 $\times$ 5 = 12
a + 25 = 12
a = -13

Q.33:

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

Solution:

Sum of first n natural numbers is:

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

$\frac{n(n+1)}{2}>1000$
$n(n+1)>2000$
Now test values:
${n}=44$ :
$44 \times 45=1980$ (not enough)
${n}=45$ :
$45 \times 46=2070$ (greater than 2000)
So smallest value is ${n}=45$

Q.34:

Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Solution:

Given GP: 2, 8, 32, …
First term:

$a=2$
Common ratio:

$r=8 / 2=4$
Explicit formula (nth term)
For a GP:

$a_n=a \cdot r^{n-1}$
So,

$a_n=2 \cdot 4^{n-1}$
Find which term is 131072

$2 \cdot 4^{n-1}=131072$
Divide by 2 :

$4^{n-1}=65536$
Now,

$65536=4^8\left(\text { since } 4^8=\left(2^2\right)^8=2^{16}=65536\right)$
So,

$n-1=8$
$n=9$
Recursive formula
First term: $a_1=2$
Recurrence relation: $a_n=4 \cdot a_{n-1}, n \geq 2$

Q.35:

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.

Solution:

Let the GP be:

$a, a r, a r^2$
Given 1: Sum of first three terms

$a+a r+a r^2=\frac{13}{12}$

\begin{equation*} a\left(1+r+r^2\right)=\frac{13}{12} \end{equation*} …(1)
Given 2: Product of three terms

$a \cdot a r \cdot a r^2=-1$
$a^3 r^3=-1$
$(a r)^3=-1$
So,

\begin{equation*} a r=-1 \end{equation*} …(2)
Step 1: Use (2) in (1)
From (2), $a=-\frac{1}{r}$
Substitute in (1):

$-\frac{1}{r}\left(1+r+r^2\right)=\frac{13}{12}$
Multiply both sides by $r$ :

$-\left(1+r+r^2\right)=\frac{13 r}{12}$
Multiply by 12 :

$-12\left(1+r+r^2\right)=13 r$
$-12-12 r-12 r^2=13 r$

Bring all terms to one side:

$-12-25 r-12 r^2=0$
Multiply by -1 :

$12 r^2+25 r+12=0$
Step 2: Solve quadratic

$12 r^2+25 r+12=0$
Discriminant:

$25^2-4 \cdot 12 \cdot 12=625-576=49$
$r=\frac{-25 \pm 7}{24}$
So:
$r=\frac{-18}{24}=-\frac{3}{4}$
$r=\frac{-32}{24}=-\frac{4}{3}$

Step 3: Find terms
Using $a r=-1$
Case 1: $r=-\frac{3}{4}$

$a=\frac{-1}{r}=\frac{-1}{-\frac{3}{4}}=\frac{4}{3}$
Terms:

$\frac{4}{3},-1, \frac{3}{4}$

Case 2: $r=-\frac{4}{3}$

$a=\frac{-1}{-\frac{4}{3}}=\frac{3}{4}$
Terms:

$\frac{3}{4},-1, \frac{4}{3}$

Q.36:

If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Solution:

Let the GP be:
${a}, {ar}, {ar}^2, {ar}^3, \ldots$
Given:
4th term $={x}={ar}^3$
10th term $={y}={ar}^9$
16th term = z = ar ${ }^{15}$

Now consider $y^2$ :

$y^2=\left(a r^9\right)^2=a^2 r^{18}$
Now consider $x z$ :

$x z=\left(a r^3\right)\left(a r^{15}\right)=a^2 r^{18}$
Compare:

$y^2=a^2 r^{18}=x z$
So,

$y^2=x z$
Conclusion:
Since $y^2=x z$, the numbers ${x}, {y}, {z}$ are in Geometric Progression (GP).
Hence proved.

Q.37:

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.

Solution:

Let the three terms of the GP be:

$a, a r, a r^2$
Given:

$a+a r+a r^2=26$ …(1)
$a^2+a^2 r^2+a^2 r^4=364$ …(2)
From (1):

$a\left(1+r+r^2\right)=26$
From (2):

$a^2\left(1+r^2+r^4\right)=364$
Now divide (2) by (1) ${ }^2$ :

$\frac{1+r^2+r^4}{\left(1+r+r^2\right)^2}=\frac{364}{26^2}$
$\frac{1+r^2+r^4}{\left(1+r+r^2\right)^2}=\frac{7}{13}$
Using identity:

$1+r^2+r^4=\left(1+r+r^2\right)^2-2 r\left(1+r+r^2\right)$
So,

$1-\frac{2 r}{1+r+r^2}=\frac{7}{13}$
$\frac{2 r}{1+r+r^2}=\frac{6}{13}$

$\frac{r}{1+r+r^2}=\frac{3}{13}$
Solving:

$13 r=3\left(1+r+r^2\right)$
$3 r^2-10 r+3=0$
Solving quadratic:

$r=3 \text { or } r=\frac{1}{3}$
Case 1: $r=3$

$a(1+3+9)=26 \Rightarrow 13 a=26 \Rightarrow a=2$
Terms:

$2,6,18$
Case 2: $r=1 / 3$

$a\left(1+\frac{1}{3}+\frac{1}{9}\right)=26 \Rightarrow a=18$
Terms:

$18,6,2$

Q.38:

Suppose $P_1=1, P_2=2$ and for $n>2, P_n=P_1+P_2+\cdots+P_{n-1}+1$. Find the values of ${P}_1, {P}_2, \ldots, {P}_8$. Can you find a simpler recursive formula for ${P}_{{n}}$? Can you give an explicit formula?

Solution:

Given:

$P_1=1, P_2=2$

and for $n>2$,

$P_n=P_1+P_2+\cdots+P_{n-1}+1$
Step 1: Find first 8 terms
We calculate step by step:

• $P_1=1$
• $P_2=2$

Now,

$P_3=1+2+1=4$
$P_4=1+2+4+1=8$
$P_5=1+2+4+8+1=16$
$P_6=32$
$P_7=64$
$P_8=128$
So, first 8 terms are:

$1,2,4,8,16,32,64,128$
Step 2: Simpler recursive formula
We observe: each term is double of previous term.

So,

$P_n=2 P_{n-1}, n \geq 2$

with $P_1=1$

Step 3: Explicit formula
This is a GP with:

• first term $a=1$
• common ratio $r=2$

So,

$P_n=2^{n-1}$
Final Answer:
First 8 terms:

$1,2,4,8,16,32,64,128$
Recursive formula:

$P_1=1, P_n=2 P_{n-1}$
Explicit formula:

$P_{{n}}=2^{n-1}$

Q.39:

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

Solution:

Given:

$W_1=1, W_2=2$

and for $n>2$,

$W_n=W_1+W_2+\cdots+W_{n-2}+2$
Step 1: Find terms one by one
We use the given rule carefully

$W_3=W_1+2=1+2=3$

$W_4=W_1+W_2+2=1+2+2=5$

$W_5=W_1+W_2+W_3+2=1+2+3+2=8$

$W_6=1+2+3+5+2=13$

$W_7=1+2+3+5+8+2=21$

$W_8=1+2+3+5+8+13+2=34$
Step 2: Write the sequence

$1,2,3,5,8,13,21,34$

Step 3: Recognition
If we observe carefully, from the 3rd term onwards:

$3=1+2, 5=2+3, 8=3+5, \ldots$
So this is the Fibonacci sequence, with a small modification in the definition.
Values are:

$1,2,3,5,8,13,21,34$
This is a form of the Fibonacci sequence.