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

Students often refer to Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions Solutions to verify their answers.

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Solutions

Class 9 Ganita Manjari Chapter 8 Solutions

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

Think and Reflect (NCERT Pages 176-177)

Question 1.
Consider the sequence 1, 4, 7, 10, 13,…. Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms upto a given term of this sequence?
Solution:
Given sequence is 1, 4, 7, 10, 13,….
Here, the first term = 1
the second term = 4 = 1 + 3
the third term = 7 = 4 + 3
the fourth term = 10 = 7 + 3
Here, the next term of the sequence is given by adding 3 to the preceding term.
Hence, the next four terms of sequence are 13 + 3 = 16, 16 + 3 = 19, 19 + 3 = 22 and 22 + 3 = 25.
The first 10 terms of this sequence are 1, 4, 7, 10, 13, 16, 19, 22, 25, 28.
The sequence obtained by adding all the terms upto a given term of this sequence is given by
First term = 1
Second term = 1 + 4 = 5
Third term = 1 + 4 + 7 = 12
Fourth term = 1 + 4 + 7 + 10 = 22
Fifth term = 1 + 4 + 7 + 10 + 13 = 35
Sixth term = 1 + 4 + 7 + 10 + 13 + 16 = 51
Seventh term = 1 + 4 + 7 + 10 + 13 + 16 + 19 = 70
Eighth term = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 = 92
Ninth term = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 = 117
Tenth term = 1 + 4 + 7 + 10 + 13+ 16 + 19 + 22 + 25 + 28 = 145
Hence, the first 10 terms are 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

Question 2.
Can you write t₅, t₆, t₇, and t₈ for the sequence of triangular numbers?
Solution:
Each term of the triangular number sequence is the sum of the natural numbers upto that term.
Hence, t₅ = 1 + 2 + 3 + 4 + 5 = 15
t₆ = 1 + 2 + 3 + 4 + 5 + 6 = 21
t₇ = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
t₈ = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Think and Reflect (NCERT Page 177)

Question 1.
Why is it useful to have an explicit formula for the nth term of a sequence?
Solution:
Having an explicit formula for the nth term of a sequence is useful because it allows you to calculate any specific term directly without listing all previous terms. It saves time and effort, especially for large values of n, and helps in identifying patterns, making calculations easier, and solving problems efficiently. Hence, an explicit formula provides a direct and quick way to find any term of the sequence.

Question 2.
Using the explicit rule u_n = 2n – 1, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.
Solution:
Given, u_n = 2n – 1
On putting n = 53, we get 53rd term
i.e. u₅₃ = 2(53) – 1 = 106 – 1 = 105
On putting n = 108, we get 108th term
i.e. u₁₀₈ = 2(108) – 1 = 216 – 1 = 215
On putting n = 1170, we get 1170th term
i.e. u₁₁₇₀ = 2(1170) – 1 = 2340 – 1 = 2339

Question 3.
Consider the sequence that is generated by the explicit formula S_n = 5n – 2. Can you write the first 6 terms of this sequence? What is the 100th term? The 1000th term?
Solution:
Given, S_n = 5n – 2
On putting n = 1, we get
S₁ = 5(1) – 2 = 5 – 2 = 3
On putting n = 2, we get
S₂ = 5(2) – 2 = 10 – 2 = 8
On putting n = 3, we get
S₃ = 5(3) – 2 = 15 – 2 = 13
On putting n = 4, we get
S₄ = 5(4) – 2 = 20 – 2 = 18
On putting n = 5, we get
S₅ = 5(5) – 2 = 25 – 2 = 23
On putting n = 6, we get
S₆ = 5(6) – 2 = 30 – 2 = 28
The first six terms of the sequence are 3, 8, 13, 18, 23, 28.
The 100th term of the sequence,
S₁₀₀ = 5(100) – 2 = 500 – 2 = 498
The 1000th term of the sequence,
S₁₀₀₀ = 5(1000) – 2 = 5000 – 2 = 4998

Question 4.
Can you find the rule describing the nth term of the sequence of square numbers?
Solution:
The sequence of square numbers is 1, 4, 9, 16, 25,……
Now, we find the pattern of the sequence.
Here, 1 = 1², 4 = 2², 9 = 3², 16 = 4², 25 = 5²,……
So, each term is the square of its position number.
Let the nth term be an.
∴ a_n = n²
Hence, the rule for the nth term is a_n = n².

Question 5.
Here is the sequence of the first ten prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Do you see any pattern in this sequence? Can you think of a rule that can predict the next few prime numbers?
Solution:
Given sequence is 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
These are prime numbers, i.e., numbers having only two factors, 1 and the number itself.
Let us observe the pattern by finding the differences
3 – 2 = 1, 5 – 3 = 2, 7 – 5 = 2, 11 – 7 = 4, 13 – 11 = 2,….
Hence, the sequence does not follow any simple pattern.
Also, unlike some other sequences, there is no simple formula to find the nth prime number.
Except for 2 and 3, all prime numbers are of the form 6n – 1 or 6n + 1.
Now, 5 = 6(1) – 1, 7 = 6(1) + 1
11 = 6(2) – 1, 13 = 6(2) + 1
17 = 6(3) – 1, 19 = 6(3) + 1
Thus, this pattern helps us identify possible prime numbers.
Checking (for next primes after 29)
31 = 6(5) + 1
35 = 6(6) – 1 (not prime)
37 = 6(6) + 1
41 = 6(7) – 1
43 = 6(7) + 1
So, not all numbers of the form 6n ± 1 are prime.
There is no exact formula to generate all prime numbers.
However, the 6n ± 1 rule helps to predict possible primes, which can then be tested.

Think and Reflect (NCERT Page 178)

Question 1.
Consider the expression t_n = 3n – 7.
(i) Find its first, second, third, 12th, 18th, and 50th terms.
(ii) Which term of the sequence is 332?
(iii) Is 557 a term of this sequence? Why or why not?
Solution:
Given, the expression t_n = 3n – 7
(i) Put n = 1 to get the first term,
t₁ = 3(1) – 7 = 3 – 7 = -4
Put n = 2 to get the second term,
t₂ = 3(2) – 7 = 6 – 7 = -1
Put n = 3 to get the third term,
t₃ = 3(3) – 7 = 9 – 7 = 2
Put n = 12 to get 12th term,
t₁₂ = 3(12) – 7 = 36 – 7 = 29
Put n = 18 to get 18th term,
t₁₈ = 3(18) – 7 = 54 – 7 = 47
Put n = 50 to get 50th term,
t₅₀ = 3(50) – 7 = 150 – 7 = 143

(ii) Let the nth term of the sequence be 332.
∴ t_n = 332
⇒ 3n – 7 = 332
⇒ 3n = 332 + 7
⇒ 3n = 339
⇒ n = 113
Hence, the 113th term of the sequence is 332.

(iii) Let 557 be the nth term of the sequence.
∴ t_n = 557
⇒ 3n – 7 = 557
⇒ 3n = 557 + 7
⇒ 3n = 564
⇒ n = 188
Since n is a natural number
Therefore, 557 is the 188th term of the sequence.

Think and Reflect (NCERT Pages 182-183)

Question 1.
Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?
(i) 2, 5, 8, 11,….
(ii) -5, -1, 3, 7,….
Solution:
(i) We have the sequence 2, 5, 8, 11,….
Now, 5 – 2 = 3, 8 – 5 = 3 and 11 – 8 = 3
Here, we observe that the difference between successive terms is the same.
Therefore, this sequence is an AP with first term, a = 2, and common difference, d = 3.
Now, the nth term of the AP,
t_n = a + (n – 1)d
= 2 + (n – 1)3 [∵ a = 2 and d = 3]
= 2 + 3n – 3
= 3n – 1
Now, we find values of tn corresponding to a few values of n.

Now, we plot the ordered pairs emerging from the table, that is, (1, 2), (2, 5), (3, 8), (4, 11), (5, 14), and draw the graph by joining the points consecutively.

We observe that the points lie on a straight line.
Hence, the graph of an AP is always a straight line.

(ii) We have sequence is -5, -1, 3, 7,…..
Now, -1 – (-5) = -1 + 5 = 4, 3 – (-1) = 3 + 1 = 4 and 7 – 3 = 4
Here, we observe that the difference between successive terms is the same.
Therefore, this sequence is an AP with first term, a = -5, and common difference, d = 4.
Now, the nth term of the AP,
t_n = a + (n – 1)d
= -5 + (n – 1)4 [∵ a = -5 and d = 4]
= -5 + 4n – 4
= -9 + 4n
Now, we find the values of an corresponding to a few values of n.

Now, we plot the ordered pairs emerging from the table, that is, (1, -5), (2, -1), (3, 3), (4, 7), (5, 11), and draw the graph by joining the points consecutively.

We observe that the points lie on a straight line.
Hence, the graph of an AP is always a straight line.

Question 2.
Using the formula t_n = a + (n – 1) × d, find the nth terms of the following progressions.
(i) \(\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, \ldots\)
(ii) 1.5, 3.5, 5.5, 7.5, …
Solution:
Given, AP is \(\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, \ldots\)
Here, first term, a = \(\frac {1}{2}\)
and common difference, d = \(\frac{5}{2}-\frac{1}{2}=\frac{4}{2}\) = 2
Now, the nth term of the AP,
\(t_n=a+(n-1) d=\frac{1}{2}+(n-1) 2=\frac{1}{2}+2 n-2\)
= \(2 n+\frac{1-4}{2}=2 n-\frac{3}{2}\)

(ii) Given, AP is 1.5, 3.5, 5.5, 7.5,….
Here, the first term, a = 1.5
and common difference, d = 3.5 – 1.5 = 2
The nth term of the AP,
t_n = a + (n – 1)d
= 1.5 + (n – 1)2
= 1.5 + 2n – 2
= 2n – 0.5

Question 3.
Find recursive rules for the following APs.
(i) 2, 5, 8, 11,….
(ii) -5, -1, 3, 7,….
(iii) \(\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, \ldots\)
(iv) 1.5, 3.5, 5.5, 7.5,….
Solution:
Recursive rule for an AP is t₁ = a, t_n = t_n-1 + d for n ≥ 2,
where a is first term and d is common difference of AP.
(i) Recursive rule for 2, 5, 8, 11,… is

Think and Reflect (NCERT Pages 184)

Question 1.
If we use the notation Sn to represent the sum of the first n natural numbers, then S_n = \(\frac{n(n+1)}{2}\). Can you use this formula to find S₂₀, S_50, or S₁₀₀₀?
Solution:
Given the sum of the first n natural numbers,

Think and Reflect (NCERT Page 185)

Question 1.
The sequence tn of triangular numbers is 1, 3, 6, 10, 15,….. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus, t_n = \(\frac{n(n+1)}{2}\). Can you use this to find the 10th, 17th, and 80th triangular numbers?
Solution:
Given the nth triangular number,
\(\begin{aligned}
& t_n=\frac{n(n+1)}{2} \\
& t_{10}=\frac{10 \times 11}{2}=5 \times 11=55 \\
& t_{17}=\frac{17 \times 18}{2}=17 \times 9=153
\end{aligned}\)
and t₈₀ = \(\frac{80 \times 81}{2}\)
= 40 × 81
= 3240
Hence, t₁₀ = 55, t₁₇ = 153 and t₈₀ = 3240.

Think and Reflect (NCERT Pages 187-188)

Question 1.
Is 1, 2, 4, 8, 16,… a geometric progression? If so, what is the common ratio?
Solution:
We have, \(\frac{2}{1}=\frac{4}{2}=\frac{8}{4}=\frac{16}{8}=2\)
Here, the ratios of consecutive terms are all the same.
Therefore, the given sequence is a GP with common ratio 2.

Question 2.
Is 1, 3, 9, 27, 81,… a GP? If so, what is the common ratio?
Solution:
We have, \(\frac{3}{1}=\frac{9}{3}=\frac{27}{9}=\frac{81}{27}=3\)
The ratios of consecutive pairs of terms are all the same.
Therefore, the given sequence is a GP with common ratio 3.

Question 3.
Is 1, -1, 1, -1, 1,… a GP? If so, what is the common ratio?
Solution:
We have, \(\frac{-1}{1}=\frac{1}{-1}=\frac{-1}{1}=\frac{1}{-1}=-1\)
The ratios of consecutive pairs of terms are all the same.
Therefore, the given sequence is a GP with common ratio -1.

Question 4.
Check whether the following sequences are geometric progressions and find their nth terms.
(i) 2, 10, 50, 250,….
(ii) \(4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, \ldots \ldots\)
(iii) \(3, \frac{-3}{2}, \frac{3}{4}, \frac{-3}{8}, \ldots \ldots\)
Solution:
(i) Given sequence is 2, 10, 50, 250,….
Here, t₁ = 2, t₂ = 10, t₃ = 50 and t₄ = 250.
Now, we find the ratios of consecutive pairs of terms.
Here, \(\frac{t_2}{t_1}=\frac{10}{2}=5\),
\(\frac{t_3}{t_2}=\frac{10}{2}=5\)
and \(\frac{t_4}{t_3}=\frac{10}{2}=5\)
The ratio of consecutive pairs of terms is a constant, i.e., 5.
Hence, the given sequence is a GP with a = 2 and r = 5.
The nth term of GP,
t_n = ar^n-1 = 2(5)^n-1

(ii) We have, sequence is \(4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, \ldots \ldots\)
Here, \(t_1=4, t_2=\frac{8}{3}, t_3=\frac{16}{9} \text { and } t_4=\frac{32}{27} .\)
Now, we find the ratios of consecutive pairs of terms.
\(\begin{aligned}
& \text { Here, } \frac{t_2}{t_1}=\frac{8}{3} \div 4=\frac{8}{3} \times \frac{1}{4}=\frac{2}{3} \\
& \frac{t_3}{t_2}=\frac{16}{9} \div \frac{8}{3}=\frac{16}{9} \times \frac{3}{8}=\frac{2}{3} \\
& \text { and } \frac{t_4}{t_3}=\frac{32}{27} \div \frac{16}{9}=\frac{32}{27} \times \frac{9}{16}=\frac{2}{3}
\end{aligned}\)
The ratio of consecutive pairs of terms is constant, i.e., \(\frac {2}{3}\)
Hence, the given sequence is a GP, with a = 4 and r = \(\frac {2}{3}\).
The nth term of GP, \(t_n=a r^{n-1}=4 \cdot\left(\frac{2}{3}\right)^{n-1}\)

(iii) We have, sequence is \(3, \frac{-3}{2}, \frac{3}{4}, \frac{-3}{8}, \ldots \ldots\)
Here, \(t_1=3, t_2=\frac{-3}{2}, t_3=\frac{3}{4} \text { and } t_4=\frac{-3}{8}\)
Now, we find the ratios of consecutive pairs of terms
\(\begin{aligned}
& \text { Here, } \frac{t_2}{t_1}=\frac{-3}{2} \div 3=\frac{-3}{2} \times \frac{1}{3}=\frac{-1}{2}, \\
& \frac{t_3}{t_2}=\frac{3}{4} \div \frac{-3}{2}=\frac{3}{4} \times \frac{2}{-3}=\frac{-1}{2} \\
& \text { and } \frac{t_4}{t_2}=\frac{-3}{8} \div \frac{3}{4}=\frac{-3}{8} \times \frac{4}{3}=\frac{-1}{2} .
\end{aligned}\)
The ratio of consecutive pairs of terms is a constant, i.e., \(\frac {-1}{2}\).
Hence, the given sequence is a GP with a = 3 and r = \(\frac {-1}{2}\).
The nth term of GP, \(t_n=a r^{n-1}=3 \cdot\left(\frac{-1}{2}\right)^{n-1}\)

Question 5.
Can you find a recursive rule for the formula \(t_n=3 \times 10^{n-1}\) that generates the geometric progression 3, 30, 300, 3000,…..?
Solution:

Ex 8.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 8.1 Solutions

Exercise 8.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 8.1 Solutions

Question 1.
Find the first five terms of the sequence in which the nth term is given by (i) t_n = 3n – 4, (ii) t_n = 2 – 5n and (iii) t_n = n² – 2n + 3 for n ≥ 1.
Solution:
(i) Given, t_n = 3n – 4, n ≥ 1
Put n = 1 to get 1st term,
t₁ = 3(1) – 4 = 3 – 4 = -1
Put n = 2 to get 2nd term,
t₂ = 3(2) – 4 = 6 – 4 = 2
Put n = 3 to get 3rd term,
t₃ = 3(3) – 4 = 9 – 4 = 5
Put n = 4 to get 4th term,
t₄ = 3(4) – 4 = 12 – 4 = 8
Put n = 5 to get 5th term,
t₅ = 3(5) – 4 = 15 – 4 = 11
Hence, the first five terms of the sequence t_n = 3n – 4 are -1, 2, 5, 8, 11.

(ii) Given, t_n = 2 – 5n, n ≥ 1
Put n = 1 to get the 1st term,
t₁ = 2 – 5(1) = 2 – 5 = -3
Put n = 2 to get 2nd term,
t₂ = 2 – 5(2) = 2 – (10) = -8
Put n = 3 to get 3rd term,
t₃ = 2 – 5(3) = 2 – 15 = -13
Put n = 4 to get 4th term,
t₄ = 2 – 5(4) = 2 – 20 = -18
Put n = 5 to get 5th term,
t₅ = 2 – 5(5) = 2 – 25 = -23
Hence, the first five terms of the sequence t_n = 2 – 5n are -3, -8, -13, -18, -23.

(iii) Given,tn = n² -2n + 3, n² ≥ 1
Put n = 1 to get 1st term,
t₁ = (1)² – 2(1) + 3 = 1 – 2 + 3 = 2
Put n = 2 to get 2nd term,
t₂ = (2)² – 2(2) + 3 = 4 – 4 + 3 = 3
Put n = 3 to get 3rd term,
t₃ = (3)² – 2(3) + 3 = 9 – 6 + 3 = 6
Put n = 4 to get 4th term,
t₄ = (4)² – 2(4) + 3 = 16 – 8 + 3 = 11
Put n = 5 to get 5th term,
t₅ = (5)² – 2(5) + 3 = 25 – 10 + 3 = 18
Hence, the first five terms of the sequence t_n = n² – 2n + 3 are 2, 3, 6, 11, 18.

Question 2.
Find the 10th and 15th terms of the sequence t_n = 5n – 3 for n ≥ 1.
Solution:
Given the sequence, t_n = 5n – 3
On putting n = 10, we get
10th term, t₁₀ = 5(10) – 3 = 50 – 3 = 47
On putting n = 15, we get
15th term, t₁₅ = 5(15) – 3 = 75 – 3 = 72

Question 3.
Determine whether 97 and 172 are terms of the sequence t_n = 5n – 3 for n ≥ 1.
Solution:
Let 97 be the nth term of the sequence t_n = 5n – 3.
∴ t_n = 97
⇒ 5n – 3 = 97
⇒ 5n = 97 + 3
⇒ 5n = 100
⇒ n = 20
Hence, the 20th term of the sequence t_n = 5n – 3 is 97.
Let 172 be the mth term of the sequence t_n = 5n – 3.
∴ t_m = 172
⇒ 5m – 3 = 172
⇒ 5m = 172 + 3
⇒ 5m = 175
⇒ m = 35
Hence, the 35th term of the sequence t_m = 5n – 3 is 172.

Question 4.
Which term of the sequence, t_n = 5n – 3 for n ≥ 1, is 607?
Solution:
Let the nth term of the sequence t_n = 5n – 3 be 607.
∴ t_n = 607
⇒ 5n – 3 = 607
⇒ 5n = 607 + 3
⇒ 5n = 610
⇒ n = 122
Hence, the 122nd term of the given sequence is 607.

Question 5.
A sequence is given by the recursive rule t₁ = -5, t_n+1 = t_n + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Solution:
Given, t₁ = -5 and t_n+1 = t_n + 3
Put n = 1 to get 2nd term,
t₁₊₁ = t₁ + 3
⇒ t₂ = -5 + 3 = -2
Put n = 2 to get 3rd term,
t₂₊₁ = t₂ + 3
⇒ t₃ = -2 + 3 = 1
Put n = 3 to get 4th term,
t₃₊₁ = t₃ + 3
⇒ t₄ = 1 + 3 = 4
Put n = 4 to get 5th term,
t₄₊₁ = t₄ + 3
⇒ t₅ = 4 + 3 = 7
Hence, the first five terms of the sequence are -5, -2, 1, 4, 7.
Here, we get each term by adding 3 to the previous term.
Thus, the nth term of the sequence can be deduced by taking the first term and adding 3 repeatedly (n – 1) times.
i.e. t_n = -5 + 3(n – 1)
= -5 + 3n – 3
= -8 + 3n
Let 52 be the nth term of this sequence.
∴ t_n = 52
⇒ -8 + 3n = 52
⇒ 3n = 52 + 8
⇒ 3n = 60
⇒ n = 20
Hence, 52 is the 20th term of the sequence.

Question 6.
Let T₁ = 1, T₂ = 2, T₃ = 4, and T_n = T_n-1 + T_n-2 + T_n-3 for n ≥ 4. Find T₄, T₅, T₆, T₇, and T₈.
Solution:

Ex 8.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 8.2 Solutions

Exercise 8.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 8.2 Solutions

Question 1.
Find the 10th and 26th terms of the AP 3, 8, 13, 18,……
Solution:
Let the first term of an AP be a₁ = 3.
The common difference of AP is d = t₂ – t₁
= 8 – 3
= 5
General term of an AP is given by t_n = a + (n – 1)d
Now, the 10th term of AP,
t₁₀ = a + (10 – 1)d
= 3 + 9(5) [∵ a = 3 and d = 5]
= 3 + 45
= 48
and the 26th term of AP
t₂₆ = a + (26 – 1)d
= a + 25d
= 3 + 25(5)
= 3 + 125
= 128 [∵ a = 3 and d = 5]
Hence, t₁₀ = 48 and t₂₆ = 128.

Question 2.
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 is 21, 18, 15,
Here, the first term, a = t₁ = 21
and common difference, d = t₂ – t₁
= 18 – 21
= -3
General term of an AP is given by t_n = a + (n – 1)d
Let -81 be the nth term of the AP.
∴ -81 = a + (n – 1)d
⇒ 81 = 21 + (n – 1)(-3) [∵ a = 21 and d = -3]
⇒ -81 – 21 = (n – 1)(-3)
⇒ -102 = (n – 1)(-3)
⇒ n – 1 = 34
⇒ n = 34 + 1
⇒ n = 35
Therefore, -81 is the 35th term of the AP.
Let 0 be the mth term of the AP.
0 = a + (m – 1)d
⇒ 0 = 21 + (m – 1)(-3) [∵ a = 21 and d = -3]
⇒ -21 = (m – 1)(-3)
⇒ m – 1 = 7
⇒ m = 7 + 1 = 8
Yes, 0 is the 8th term of the AP.

Question 3.
Find the nth term of the AP: 11, 8, 5, 2,…. Write the recursive rule for this AP.
Solution:
Given the AP is 11, 8, 5, 2,…
Here, first term (a) = t₁ = 11
and common difference, (d) = t₂ – t₁ = 8 – 11 = -3.
Now, the nth term of an AP, t_n = a + (n – 1)d
⇒ t_n = 11 + (n – 1)(-3) [∵ a = 11 and d = -3]
= 11 + (-3n) + 3
= 14 – 3n
For the AP is 11, 8, 5, 2,…. we are subtracting 3 from each term to get the next term.
So, the recursive rule for the AP is t_n+1 + 1 = t_n – 3, t₁ = 11 for n > 1.

Question 4.
An AP consists of 50 terms in which the 3rd term is 12, and the last term is 106. Find the 29th term.
Solution:
Given, t₃ = 12 and t₅₀ = 106 [∵ the last term will be 50th term]
The nth term of AP is given by
t_n = a + (n – 1)d
⇒ t₃ = a + (3 – 1)d
⇒ 12 = a + 2d …..(i)
and t50 = a + (50 – 1)d
106 = a + 49d …..(ii)
On subtracting Eq. (i) from Eq. (ii), we get
106 – 12 = a + 49d – a – 2d
⇒ 94 = 47d
⇒ d = 47
On putting d = 2 in Eq. (i), we get
12 = a + 2(2)
⇒ 12 = a + 4
⇒ 12 – 4 = a
⇒ a = 8
The 29th term of AP,
t₂₉ = 8 + (29 – 1)2
= 8 + 28 × 2
= 8 + 56
= 64

Question 5.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution:
The list of 2-digit numbers, which are divisible by 3, is 12, 15, 18, 21,…, 99.
Here, 15 – 12 = 3, 18 – 15 = 3 and 21 – 18 = 3
Hence, the list of numbers is an AP with a = 12 and d = 3.
Let 99 be the nth term of an AP.
∴ t_n = a + (n – 1)d
⇒ 99 = 12 + (n – 1) (3)
⇒ 99 – 12 = (n – 1)(3)
⇒ 87 = 3(n – 1)
⇒ n – 1 = 29
⇒ n = 30
Hence, the 2-digit number which are divisible by 3 are 30.
Now, the sum of all 2-digit numbers that are divisible by 3.
12 + 15 + 18 + 21 +… + 99 = 3[4 + 5 + 6 +… + 33]
= 3[1 + 2 + 3] + 3[4 + 5 +… + 33] – 3[1 + 2 + 3]
= 3[1 + 2 + 3 + … + 33] – 3[1 + 2 + 3]
= \(3\left[\frac{33 \times(33+1)}{2}\right]-3\left[\frac{3 \times(3+1)}{2}\right]\) [∵ Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)]
= \(3\left[\frac{33 \times 34}{2}\right]-3\left[\frac{3 \times 4}{2}\right]\)
= 3[33 × 17] – 3[3 × 2]
= 3[561] – 3[6]
= 1683 – 18
= 1665
Hence, the sum of all 2-digit numbers which are divisible by 3 is 1665.

Question 6.
Harish started work at an annual salary of ₹ 500000 and received an increment of ₹ 20000 each year. After how many years did his income reach ₹ 700000?
Solution:
Given, Harish’s annual salary = ₹ 500000
and salary increment each year = ₹ 20000.
Since each year’s salary gets increased by a fixed amount.
So, this situation forms an AP, with a1 = 500000 and d = 20000.
Let his salary reach ₹ 700000 after n yr.
∴ t_n = 700000
⇒ a + (n – 1)d = 700000
⇒ 500000 + (n – 1) (20000) = 700000
⇒ (n – 1) (20000) = 700000 – 500000
⇒ (n – 1) (20000) = 200000
⇒ n – 1 = 10
⇒ n = 11
Hence, the salary becomes ₹ 700000 in the 11th year.

Question 7.
A child arranges marbles in rows so that the first row has 1 marble, the second row has 2 marbles, the third row has 3, and so on up to 25 rows. How many marbles does the child use in all?
Solution:
Given,
Marbles in 1st row = 1
Marbles in 2nd row = 2
Marbles in 3rd row = 3
Marbles in 25th row = 25
∴ Sum of all marbles = 1 + 2 + 3 + 4 + …. + 25
= \(\frac{25(25+1)}{2}\) [∵ sum of first n natural numbers = \(\frac{n(n+1)}{2}\)]
= \(\frac{25 \times 26}{2}\)
= 25 × 13
= 325

Ex 8.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 8.3 Solutions

Exercise 8.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 8.3 Solutions

Question 1.
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution:
The nth term of a GP is given by \(t_n=a r^{n-1}\), where a is the first term and r is the common ratio.
Here, t₈ = 192 and r = 2

Question 2.
Find the 10th and nth terms of the GP 5, 25, 125,….
Solution:
Given, GP is 5, 25, 125,….
Here, the first term (a) = 5
and common ratio (r) = \(\frac {25}{5}\) = 5
Now, the 10th term of GP, t₁₀ = ar^10-1
= ar⁹
= 5(5)⁹
= (5)¹⁰
= 5¹⁰
The nth term of GP, t_n = ar^n-1
= 5(5)^n-1
= 5ⁿ

Question 3.
A sequence is given by the recursive rule t₁ = 2, t_n+1 = 3t_n – 2 for n ≥ 1. Which term of the sequence is 730?
Solution:
Here, t₁ = 2
and t_n+1 = 3t_n – 2 …..(i)
On putting n = 1 in Eq. (i), we get
t₂ = 3t₁ – 2
= 3(2) – 2 [∵ t₁ = 2]
= 6 – 2
= 4
On putting n = 2 in Eq. (i), we get
t₂₊₁ = 3t₂ – 2
⇒ t₃ = 3t₂ – 2
= 3(4) – 2
= 12 – 2
= 10
On putting n = 3 in Eq. (i), we get
t₃₊₁ = 3t₃ – 2
⇒ t₄ = 3(10) – 2
= 30 – 2
= 28
On putting n = 4 in Eq. (i), we get
t₄₊₁ = 3t₄ – 2
⇒ t₅ = 3(28) – 2
= 84 – 2
= 82
On putting n = 5 in Eq. (i), we get
t₅₊₁ = 3t₅ – 2
⇒ t₆ = 3(82) – 2
= 246 – 2
= 244
On putting n = 6 in Eq. (i), we get
t₆₊₁ = 3t₆ – 2
⇒ t₇ = 3(244) – 2
= 732 – 2
= 730
Hence, 730 is the 7th term of the sequence.

Question 4.
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:
We have, first term (a) of GP is 2 and common ratio (r) = \(\frac {6}{2}\) = 3
Let 4374 be the nth term of the GP.
\(\begin{array}{ll}
\therefore & 4374=a r^{n-1} \Rightarrow 4374=2(3)^{n-1} \\
\Rightarrow & 3^{n-1}=\frac{4374}{2} \Rightarrow 3^{n-1}=2187 \\
\Rightarrow & 3^{n-1}=3^7 \Rightarrow n-1=7 \Rightarrow n=8
\end{array}\)
Hence, 4374 is the 8th term of the GP.
Explicit formula for the nth term is \(t_n=2 \times 3^{n-1}\).
Recursive formula for the n th term,
t_n = 3t_n-1, t₁ = 2 for n ≥ 1

Question 5.
A ball is dropped from a height of 80 m. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues to bounce in this way, each time rising to 60% of the previous height.
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Solution:
Initial height of ball = 80 m
Each bounce reaches 60% of the previous height.
This means each new height is obtained by multiplying the previous height by a constant factor \(\frac {60}{100}\) i.e. \(\frac {3}{5}\).
Hence, the sequence of heights after bounces is \(80 \times \frac{3}{5}, 80 \times\left(\frac{3}{5}\right)^2, 80 \times\left(\frac{3}{5}\right)^3, \ldots \ldots \ldots .\)
This is a GP with a = 80 × \(\frac {3}{5}\) and r = \(\frac {3}{5}\).
(i) The height after the 5th bounce,
\(t_5=80 \times \frac{3}{5} \times\left(\frac{3}{5}\right)^{5-1} \quad\left[\because t_n=a r^{n-1}\right]\)
= \(80 \times\left(\frac{3}{5}\right)^5\)
= 80 × (0.6)5
= 80 × 0.07776
= 62208

(ii) Total distance travelled by the ball till the 6th ground hit = Initial height + Twice the sum of height of first five bounces
= \(80+2\left[80 \times \frac{3}{5}+80 \times\left(\frac{3}{5}\right)^2 \ldots+80 \times\left(\frac{3}{5}\right)^5\right]\)
= \(80+2 \times 80\left[\frac{3}{5}+\left(\frac{3}{5}\right)^2+\left(\frac{3}{5}\right)^3+\left(\frac{3}{5}\right)^4+\left(\frac{3}{5}\right)^5\right]\)
= 80 + 160[0.6 + 0.36 + 0216 + 0.1296 + 0.07776]
= 80 + 160 × 1.38336
= 80 + 221.3376
= 301.3376 m

Question 6.
Which term of the sequence 2, 2√2, 4,… is 128?
Solution:
We have, sequence is 2, 2√2, 4,…
Here, \(\frac{2 \sqrt{2}}{2}=\frac{4}{2 \sqrt{2}}=\sqrt{2}\)
The ratio of consecutive pairs of terms is a constant, √2.
Hence, the sequence is a GP with a = 2 and r = √2.
Let 128 be the nth term of the GP.
∴ 128 = ar^n-1
⇒ 128 = \(2(\sqrt{2})^{n-1}\)
⇒ \((\sqrt{2})^{n-1}\) = 64
⇒ \(2^{\frac{n-1}{2}}=2^6\)
⇒ \(\frac{n-1}{2}\) = 6
⇒ n – 1 = 12
⇒ n = 13
Hence, the 13th term of the GP is 128.

Question 7.
The given figure shows Stages 0 to 3 of the Sierpinski square carpet. Stage 0 of the 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.
(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 nth 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 region at the nth stage. What happens to this area as n, the number of states, goes on increasing?
Solution:
(i) Red squares in Stage 0 = 1
Red squares in State 1 = 8
Red squares in Stage 2 = 8² = 64
Red square in Stage3 = 8³ = 512

(ii) Red squares in Stage 4 = 8⁴ = 4096
Red squares in Stage 5 = 8⁵ = 32768

(iii) Number of red squares at stage n.
∴ t_n = 8ⁿ
The explicit formula is t_n = 8ⁿ
Recursive formula is t₀ = 1, t_n = 8t_n-1 for n ≥ 1.

(iv) Since, \(\frac {1}{9}\) part of square is remove at each stage.
Area at Stage 0 = 1 sq unit
Area at State 1 = 1 – \(\frac {1}{9}\) = \(\frac {8}{9}\) sq unit
Area at Stage 2 = \(\frac{8}{9}-\frac{1}{9}\left(\frac{8}{9}\right)^1=\frac{72-8}{81}\)
= \(\frac{64}{81}=\left(\frac{8}{9}\right)^2 \text { sq unit }\) sq unit
Area at Stage 3 = \(\left(\frac{8}{9}\right)^2-\frac{1}{9}\left(\frac{8}{9}\right)^2=\frac{512}{729}=\left(\frac{8}{9}\right)^3\) squnit
Following the pattern, we find that
Area at nth Stage = \(\left(\frac{8}{9}\right)^n\) sq unit
Area at Stage 4 = \(\left(\frac{8}{9}\right)^4\) sq unit
Area at Stage 5 = \(\left(\frac{8}{9}\right)^5\) sq unit
Recursive formula is t0 = 1, \(t_n=\frac{8}{9} t_{n-1}\) for n ≥ 1.
As the number of stages increases, the area of the red region approaches zero, but never zero.

Ganita Manjari Class 9 Maths Chapter 8 End of Chapter Exercises Solutions

Predicting What Comes Next Exploring Sequences and Progressions End of Chapter Exercises Solutions

Question 1.
Find the 31st term of an AP, whose 11th term is 38 and 16th term is 73.
Solution:
Let an be the nth term of an AP.
Here, a₁₁ = 38 and a₁₆ = 73
We know that an = a + (n – 1)d, where a is the first term and d is the common difference of the AP.
∴ a + 10d = 38 …..(i)
and a + 15d = 73 ….(ii)
On subtracting Eq. (i) from Eq. (ii), we get
5d = 35
⇒ d = 7
On putting the value of d in Eq. (i), we get
a + 10(7) = 38
⇒ a = 38 – 70
⇒ a = -32
Now, a₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Hence, the 31st term is 178.

Question 2.
Determine the AP, whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Solution:
Let tn be the nth term of an AP.
t₃ = 16 and t₇ – t₅ = 12
Again, let a be the first term and d be the common difference of the AP.
Now, t₃ = 16
⇒ a + 2d = 16 …..(i)
and t₇ – t₅ = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
From Eq. (i),
a + 2(6) = 16
⇒ a + 12 = 16
⇒ a = 4
So, the AP is a, a + d, a + 2d, a + 3d,….
i.e. 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6,…..
Thus, the AP is 4, 10, 16, 22, 28,…..

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]
Solution:
The 3-digit numbers divisible by 7 are 105, 112,…, 994.
Since the difference between any two successive terms is the same.
Therefore, the numbers are in AP with a = 105 and d = 112 – 105 = 7.
Let 994 be the nth term.
∴ a + (n – 1)d = 994
⇒ 105 + (n – 1)7 = 994
⇒ (n – 1)7 = 994 – 105
⇒ (n – 1) × 7 = 889
⇒ n – 1 = 127
⇒ n = 128
Hence, 3-digit numbers, which are divisible by 7, are 128.

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.]
Solution:
The list of multiples of 4 between 10 and 250 is 12, 16, 20,… 248.
Since the difference between any two successive terms is the same.
Therefore, this forms an AP with a = 12 and d = 16 – 12 = 4.
Let 248 be the nth term of AP.
∴ a + (n – 1)d = 248
⇒ 12 + (n – 1)4 = 248
⇒ (n – 1)4 = 248 – 12
⇒ 4(n – 1) = 236
⇒ (n – 1) = 59
⇒ n = 59 + 1
⇒ n = 60
Hence, there are 60 multiples of 4 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.
Solution:
Let a be the first term and r be the common ratio of a GP. Given, sum of first two terms = -4
⇒ a + ar = -4
⇒ a(1 + r) = -4 …..(i)
Also, t₅ = 4t₃ [given]
⇒ ar⁴ = 4ar²
⇒ r² = 4
⇒ r = ±2
Case I: When r = 2
On putting the value of r in Eq. (i), we get
a(1 + 2) = -4
a = \(\frac {-4}{3}\)
Now, the GPs are a, ar, ar² ….
i.e. \(\frac{-4}{3}, \frac{-4}{3} \times 2, \frac{-4}{3} \times(2)^2, \ldots .\)
Therefore, GP is \(\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots \ldots .\)
Case II: When r = -2
On putting the value of r in Eq. (i), we get
a(1 – 2) = -4
⇒ -a = -4
⇒ a = 4
Now, the GPs are a, ar, ar² …
i.e. 4, 4 × (-2), 4(-2)²,…
Therefore, GP is 4, -8, 16, -32,…..

Question 6.
Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Solution:
Let 100 be the sum of consecutive natural numbers from m to n, where m < n.
∴ 100 = m + (m + 1) + (m + 2) +… n
Since the sum of the first n natural numbers,
1 + 2 + ….. + n = \(\frac{n(n+1)}{2}\)
∴ 100 = \(\frac{n(n+1)}{2}-\frac{m(m+1)}{2}\)
⇒ 200 = n² + n – m² – m
⇒ 200 = n² – m² + n – m
⇒ 200 = (n + m)(n – m) + (n – m)
⇒ 200 = (n – m)(n + m + 1)
Now, take factor pairs of 200: (1, 200), (2, 100), (4, 50), (5, 40), (8, 25), (10, 20).
Case I: When n – m = 1 and n + m + 1 = 200
⇒ m + 1 + m + 1 = 200
⇒ 2m + 2 = 200
⇒ 2m = 198
⇒ m = 99
∴ n = m + 1 = 100
which makes the sum as
99 + 100 = 199 ≠ 200
Therefore, this case is nth possible.
Case II: n – m = 2 and n + m + 1 = 100
⇒ n = m + 2
⇒ m + 2 + m + 1 = 100
⇒ 2m = 100 – 3
⇒ 2m = 97
⇒ m = \(\frac {97}{2}\)
Which can’t be true as m is an integer.
Case III: n – m = 4 and n + m + 1 = 50
n = 4 + m
⇒ 4 + m + m + 1 = 50
⇒ 2m + 5 = 50
⇒ 2m = 45
⇒ m = \(\frac {45}{2}\)
Which can’t be true, as m is an integer.
Case IV: n – m = 5 and n + m + 1 = 40
⇒ n = m + 5
m + 5 + m + 1 = 40
⇒ 2m + 6 = 40
⇒ 2m = 34
⇒ m = 17
and n = 17 + 5 = 22
Therefore, the sequence is 18 + 19 + 20 + 21 + 22 = 100.
Case V: n – m = 8 and n + m + 1 = 25
⇒ n = m + 8
⇒ m + 8 + m + 1 = 25
⇒ 2m + 9 = 25
⇒ 2m = 16
⇒ m = 8
⇒ n = 8 + 8 = 16
Therefore, the sequence is 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.
Case VI: n – m = 10 and n + m + 1 = 20
⇒ n = 10 + m
⇒ 10 + m + m + 1 = 20
⇒ 2m + 11 = 20
⇒ 2m = 9
⇒ m = \(\frac {9}{2}\)
which is not true as m is an integer.
Therefore, there are only two possible cases
i.e. 18 + 19 + 20 + 21 + 22 = 100 and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.

Question 7.
The number of bacteria in a certain culture doubles every hour. If there are 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:
Given, initial number of bacteria = 30
Since the number of bacteria doubles every hour, it forms a GP with a = 30 and r = 2.
Now, nth term of GP, t_n = ar^n-1 = 30 × 2^n-1
After the 2nd hour, the number of bacteria present,
t₃ = 30 × 2^3-1 = 30 × 4 = 120
After the 4th hour, the number of bacteria present,
t₅ = 30 × 2^5-1 = 30 × 24 = 30 × 16 = 480
After nth hour, the number of bacteria present,
\(t_{n+1}=30 \times 2^{n+1-1}=30 \times 2^n\)

Question 8.
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 a be the first term of AP and d be the common difference.
Given, t₄ + t₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24 ……(i)
Also, t₆ + t₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44 ….. (ii)
On subtracting Eq. (i) from Eq.(ii), we get
2a + 14d – 2a – 10d = 44 – 24
⇒ 4d = 20
⇒ d = 5
On putting the value of d in Eq. (i), we get
2a + 10(5) = 24
⇒ 2a + 50 = 24
⇒ 2a = 24 – 50
⇒ 2a = -26
⇒ a = -13
Hence, the first three terms of AP are a, a + d, a + 2d
i.e. -13, -13 + 5, -13 + 2 × 5
i.e., -13, -8, -3.

Question 9.
Find the smallest value of n such that the sum of the first n natural numbers is greater than 1000.
Solution:
We know that the sum of the first n natural numbers is \(\frac{n(n+1)}{2}\).
For smallest n, \(\frac{n(n+1)}{2}\) > 1000
⇒ n(n + 1) > 2000
The two consecutive numbers, whose product is close to 2000, are 44 and 45, and 45 and 46.
Now, 44 × 45 = 1980 < 2000 and 45 × 46 = 2070 > 2000
Hence, the smallest value of n is 45.

Question 10.
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:
Let the first term of the GP be a and the common ratio be r.
Given, a = 2 and r = \(\frac {8}{2}\) = 4
Let 131072 be the nth term of the GP.
∴ ar^n-1 = 131072
⇒ 2 . 4^n-1 = 131072
⇒ 4^n-1 = 65536
⇒ 4^n-1 = 48
⇒ n – 1 = 8
⇒ n = 9
Therefore, 131072 is the 9th term of the GP.
Explicit formula for GP is t_n = 2 × 4^n-1
Recursive formula is a = 2, t_n = 4t^n-1 for n ≥ 2.

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.
Solution:
Let the three terms of GP be \(\frac {a}{r}\), a, ar.
where a is the first term and r is the common ratio.
The sum of the terms is

Question 12.
If the 4th, 10th, and 16th terms of a GP are x, y, z, respectively, prove that x, y, z are in GP.
Solution:
Let the first term of the GP be a and the common ratio be r.
∴ x = t₄ = ar³, \(\left[\because t_n=a r^{n-1}\right]\) …(i)
y = t₁₀ = ar⁹ ….(ii)
and z = t₁₆ = ar¹⁵ …..(iii)
To prove x, y, z are in GP
i.e., \(\frac{y}{x}=\frac{z}{y}\)
y² = xz
⇒ a²r¹⁸ = ar³ × ar¹⁵
⇒ a²r¹⁸ = a²r¹⁸ [∵ from Eqs. (i), (ii), and (iii)]
which is true.
Hence, x, y, z are in GP.

Question 13.
The sum of the first three terms of a GP is 26, and the sum of their squares is 364. Find the term of the GP.
Solution:
Let the first three terms of GP be a, ar, ar².
Given, sum of the terms = 26
⇒ a + ar + ar² = 26
⇒ a[1 + r + r²] = 26 …..(i)
⇒ a2[1 + r + r²]² = (26)² [∵ squaring both sides]
Also, the sum of squares of terms = 364
⇒ a² + (ar)² + (ar²)² = 364
⇒ a²[1 + r² + r⁴] = 364 …..(ii)
On subtracting Eq. (i) from Eq. (ii), we get
a²[1 + r² + r⁴] – a²[1 + r + r²]² = 364 – 262
⇒ a²[1 + r² + r⁴] – a²[1 + r² + r⁴ + 2r + 2r³ + 2r²] = 364 – 676
⇒ a²[1 + r² + r⁴ – 1 – r² – r⁴ – 2r – 2r³ – 2r²] = -312
⇒ a²[-2r – 2r³ – 2r²] = -312
⇒ a² × (-2r)[1 + r² + r] = -312
⇒ a²r [1 + r + r²] = 156 ……(iii)
On dividing Eq. (iii) by Eq. (i), we get
\(\frac{a^2 r\left[1+r+r^2\right]}{a\left[1+r+r^2\right]}=\frac{156}{26}\)
⇒ ar = 6
⇒ a = \(\frac {6}{r}\)
On substituting the value of a in Eq. (i), we get
\(\frac {6}{r}\)[1 + r + r²] = 26
⇒ 6 + 6r + 6r² = 26r
⇒ 6r² + 6r – 26r + 6 = 0
⇒ 6r² – 20r + 6 = 0
⇒ 3r² – 10r + 3 = 0
⇒ 3r² – 9r – r + 3 = 0
⇒ 3r(r – 3) – 1(r – 3) = 0
⇒ (r – 3)(3r – 1) = 0
⇒ r = 3 or r = \(\frac {1}{3}\)
When r = 3 then a = \(\frac {6}{r}\) = \(\frac {6}{3}\) = 2
So, the terms of GP are 2, 6, 18.
When r = \(\frac {1}{3}\) then a = \(\frac {6}{r}\)
= \(\frac{6}{1 / 3}\)
= 6 × 3
= 18
So, the terms of GP are 18, 6, 2.

Question 14.
Suppose P₁ = 1, P₂ = 2 and for n > 2, P_n = P₁ + P₂ + …. + P_n-1 + 1. Find the values of P₁, P₂,…, P₈. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?
Solution:
Here P₁ = 1
P₂ = 2
P₃ = P₁ + P₂ + 1
= 1 + 2 + 1
= 4
P₄ = P₁ + P₂ + P₃ + 1
= 1 + 2 + 4 + 1
= 8
P₅ = P₁ + P₂ + P₃ + P₄ + 1
= 1 + 2 + 4 + 8 + 1
= 16
P₆ = P₁ + P₂ + P₃ + P₄ + P₅ + 1
= 1 + 2 + 4 + 8 + 16 + 1
= 32
P₇ = P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + 1
= 1 + 2 + 4 + 8 + 16 + 32 + 1
= 64
P₈ = P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + 1
= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1
= 128
The first 8 terms of the sequence are 1, 2, 4, 8, 16, 32, 64, 128.
which is a GP with a = 1 and r = 2.
The nth term of GP is P_n = \(1 \times 2^{n-1}=2^{n-1} . \quad\left[\because t_n=a r^{n-1}\right]\)
∴ Explicit formula is P_n-1 = 2^n-1
\(\because \quad P_n=2^{n-1}=2 \times 2^{n-2}\)
Then P₁ = 1, P_n = 2 × P_n-1 for n ≥ 2.
which is a recursive formula.

Question 15.
Suppose W₁ = 1, W₂ = 2 and for n > 2, W_n = W₁ + W₂ + ….. + W_n-2 + 2. Find the values of W₁, W₂, W₃, ….., W₈. Do you recognise this sequence?
Solution:
Given, W₁ = 1, W₂ = 2 and W_n = W₁ + W₂ +…. + W_n-2 + 2
∴ W₃ = W₁ + 2
= 1 + 2
= 3,
W₄ = W₁ + W₂ + 2
= 1 + 2 + 2
= 5
W₅ = W₁ + W₂ + W₃ + 2
= 1 + 2 + 3 + 2
= 8
W₆ = W₁ + W₂ + W₃ + W₄ + 2
= 1 + 2 + 3 + 5 + 2
= 8
W₇ = W₁ + W₂ + W₃ + W₄ + W₅ + 2
= 1 + 2 + 3 + 5 + 8 + 2
= 21
and W₈ = W₁ + W₂ + W₃ + W₄ + W₅ + W₆ + 2
= 1 + 2 + 3 + 5 + 8 + 13 + 2
= 34
The sequence is 1, 2, 3, 5, 8, 13, 21, 34,……
This sequence is the Virahanka Fibonacci sequence.