NCERT Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions

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NCERT Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions

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Predicting What Comes Next Exploring Sequences and Progressions

Chapter – 8

Exercise Set 8.1

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.

Ans: (i) tn = 3n – 4,

Substitute 1, 2, 3, 4, and 5 for zz in the given equation.

t₁ = 3 × 1 – 4 = 3 – 4 = -1

t₂ = 3 × 2 – 4 = 6 – 4 = 2

t₃ = 3 × 3 – 4 = 9 – 4 = 5

t₄ = 3 × 4 – 4 = 12 – 4 = 8

t₅ = 3 × 5 – 4 = 15 – 4= 11

Thus, first five terms of the given sequence are: -1, 2, 5, 8, 11

2. Find the 10^th and 15^th terms of the sequence t_n = 5n – 3 for n ≥ 1.

Ans: Given: tn = 5n – 3

Find: 10th term

t₁₀ = 5(10) – 3 = 50 – 3 = 47

15^th term

t₁₅ = 5(15) – 3 = 75 – 3 = 72

10^th term = 47

15^th term = 72

3. Determine whether 97 and 172 are terms of the sequence t_n = 5n – 3 for n ≥ 1.

Ans: Given: t_n = 5n – 3

Substitute 97 for tn

97 = 5n – 3

⇒ 100 = 5n

⇒ n = 20

Thus, 97 is the 20th term of the given sequence.

Now, substitute t_n – 172

∴ 172 = 5n – 3

⇒ 175 = 5n

⇒ n = 35

Thus, 172 is the 35th term of the given sequence.

4. Which term of the sequence t_n = 5n – 3 for n ≥ 1 is 607?

Ans: Given: t_n = 5n – 3

We need to find n such that the term is 607.

Set:

5n – 3 = 607

5n = 610

n = 122

607 is the 122nd term of the sequence.

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?

Ans: Given: t_n = -5, t_n+1 = t_n + 3

First five terms of the sequence:

t₁ = -5

t₂ = t₁ + 3 = -5 + 3 = -2

t₃ = t₂ + 3 = -2 + 3 = 1

t₄ = t₃ + 3 = 1 + 3 = 4

t₅ = t₄ + 3 = 4 + 3 = 7

Thus, first five terms are: -5, -2, 1, 4, 7

Now, check whether 52 is a term of this sequence:

General form: tn = -5 + (n – 1).3

Substitute 52 for t_n

52 = -5 + (n – 1) × 3

⇒ 57 = 3 (n – 1)

⇒ n – 1 = 19

⇒ n = 20

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

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.

Ans: Given:

T₁ = 1, T₂ = 2, T₃ = 4

and T_n = T_n-1 + T_n-2 + T_n-3

Now find terms step by step:

T₄ = T₃ + T₂ + T₁ = 4 + 2 + 1 = 7

T₅ = T₄ + T₃ + T₂ = 7 + 4 + 2 = 13

T₆ = T₅ + T₄ + T₃ = 13 + 7 + 4 = 24

T₇ = T₆ + T₅ + T₄ = 24 + 13 + 7 = 44

T₈ = T₇ + T₆ + T₅ = 44 + 24 + 13 = 81

T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81

Exercise Set 8.2

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

Ans: The given arithmetic progression (AP) is: 3, 8, 13, 18,…

Here, the first term (a) = 3

and the common difference (d) = 8 – 3 = 5.

The formula for the 77th term of an AP is given by: tn = a + (n – 1).d

For 10th term:

Substitute n = 10. a = 3, and d = 5 into the formula:

t₁₀ = 3 + (10 – 1) × 5

= 3 + 9 × 5

= 3 + 45

= 48

Thus, the 10,h term is 48.

For 26th term:

Substitute n = 26, a = 3, and d = 5 into the formula:

t₂₆ = 3 + (26 – 1) × 5

= 3 + 25 × 5

= 3 + 125

= 128

Thus, the 26th term is 128.

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

Ans: Given AP: 21, 18, 15,…

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 – 3n + 3 = -81

24 – 3n = -81

-3n = -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 – 3n = 0

3n = 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.

3. Find the n^th term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Ans: The given AP is: 11, 8, 5, 2,…

Here, a = 11 and d = 8 – 11 = -3.

The formula for the nth term of an AP is given by: t„ = a + (n – 1).d

Substitute a = 11 and d = -3 into the formula:

t_n = 11 + (n – 1) × (-3)

= 11 – 3n + 3

= 14 – 3n

Thus, the 77th term of the AP is:

t_n = 14 – 3n

⇒ t₁ = 11

t₂ = 14 – 6 = 8 = 11 – 3

t₃ = 14 – 9 = 5 = 8 – 3

∴ The recursive rule for this AP is: t₁ = 11

t_n = t_n – 1 + (-3) = t_n-1 – 3 for n ≥ 2.

4. 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. (Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)

Ans: Given:

t₃ = a + 2d = 12

t₅₀ = a + 49d = 106

Subtract the first equation from the second:

(a + 49d) – (a + 2d) = 106 – 12

47d = 94 ⇒ d = 2

Now substitute d = 2 in a + 2d = 12:

a + 4 = 12 ⇒ a = 8

Now find the 29th term:

t₂₉ = a + (29 – 1)d = 8 + 28 × 2 = 8 + 56 = 64

29th term = 64

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

Ans: The 2-digit numbers divisible by 3 form an arithmetic progression, where the first 2-digit number divisible by 3 is 21. And the sequence is: 12, 15, 18,…, 99.

The first tenn (a) = 12,

the last term (f) = 99 and

common difference (d) = 3.

We use the formula for the nth term of an AP:

tn = a + (n – 1 )d

Substitute tn = 99, a = 12, and d = 3 into the formula:

99 = 12 + (n – 1) × 3

⇒ 99 – 12 = (n – 1) × 3

⇒ 87 = (n – 1) × 3

⇒ n – 1 = 87/3 = 29

⇒ n = 29 + 1 = 30

Thus, there are 30 terms, i.e., numbers.

Since there are 30 terms in the sequence, so. 15 pairs can be made.

The sum of each pair is (12 + 99) + (15 + 96) + (18 + 93) + …

Each pair sum to 111, and there are 15 such pairs.

Thus, the total sum is: 15 × 111 = 1665

Therefore, the sum of all the 2-digit numbers divisible by 3 is 1665.

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?

Ans: Harish’s salary forms an AP:

First term a = 5,00,000

Common difference d = 20,000

Required term = 7,00,000

Use tn = a + (n – 1)d:

7,00,000 = 5,00,000 + (n – 1) × 20,000

7,00,000 – 5,00,000 = (n – 1) × 20,000

2,00,000 = (n – 1) × 20,000

n – 1 = 10 ⇒ 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.

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?

Ans: The number of marbles in each row forms an arithmetic progression.

The first term (a) = 1, the common difference (d) = 1, number of terms (n) = 25

So the arithmetic progression is: 1 + 2 + 3 + …+ 25

This is the sum of first 25 natural numbers.

The sum of the first n natural number is given by: n(n + 1)/2

Substitute n = 25,

S₂₅ = 25(25 + 1)/2 = 25 × 26/2 = 325

Thus, the child uses 325 marbles in all.

Exercise Set 8.3

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

Ans: For a GP,

t_n = ar^n-1

Given:

Common ratio r = 2

t₈ = 192

t₈ = ar⁷ = a⋅2⁷ = 128a = 192

a = 192/128 = 3/2

Now find 12th term:

t₁₂ = ar¹¹ = 3/2⋅2¹¹ = 3/2⋅2048 = 3⋅1024 = 3072

12th term = 3072

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

Ans: Given GP: 5, 25, 125,…

We know that nth term of GP is: tn = a × r^(n-1)

For 10th term, substitute n = 10 in the above formula.

t₁₀ = 5 × 5^10-1 = 5 × 59 = 9765625 [∵ a = 5, r = 5]

The n^th term is given by: t_n = 5 × 5^n-1 = 5ⁿ

Thus, the n^th term is 5ⁿ.

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?

Ans: Given:

t₁ = 2, t_n+1 = 3t_n -2

Find terms until we reach 730:

t₁ = 2

t₂ = 3(2) – 2 = 4

t₃ = 3(4) – 2 = 10

t₄ = 3(10) – 2 = 28

t₅ = 3(28) – 2 = 82

t₆ = 3(82) – 2 = 244

t₇ = 3(244) – 2 = 730

730 is the 7th term of the sequence.

4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the 17th term.

Ans: The given geometric progression (GP) is: 2, 6, 18,…

Here, the first term (a) = 2, common ratio (r) = 3.

The nth term formula for a geometric progression is: t_n = a × r^(n-1)

We need to find the term t_n = 4374.

Substitute a = 2 and r = 3 into the formula:

4374 = 2 × 3^n-1

⇒ 2187 = 3^n-1

⇒ 37 = 3^n-1

⇒ n – 1 = 7

⇒ n = 8

Thus, the term 4374 is the 8th term of the given GP.

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^th bounce?

(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6^th time?

Ans: Given:

Initial height = 80 m

Each bounce reaches 60% = 0.6 of previous height

(i) Height after 5th bounce Heights form a GP:

80 × (0.6)ⁿ

After 5th bounce:

h₅ = 80 × (0.6)⁵ = 80 × 0.07776 = 6.2208 m

So, height ≈ 6.22 m

(ii) 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: 

h₁ = 48, h₂ = 28.8, h₃ = 17.28, h₄ = 10.368, h₅ = 6.2208

Total distance:

Distance = 80 + 2(h₁ + h₂ + h₃ + h₄ + h₅)

= 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)

= 80 + 2(110.6688) = 80+221.3376 = 301.3376 m

6. Which term of the sequence 2, 2√2, 4, … is 128?

Ans: Given sequence: 2, 2√2, 4,…

Here, first term (a) = 2 and

common ratio (r) = 2√2/2 = 4/2√2 = √2

Therefore, the nth term of GP: tn = a × r^(n-1)

Now, t_n = 128

⇒ 128 = 2 × (√2)^n-1

⇒ 64 = (√2)^n-1

⇒ (√2)¹² = (√2)ⁿ⁻¹

⇒ n – 1 = 12

⇒ n = 13

Thus, the term 128 is the 13th term of the sequence.

7. Fig. 8.12 shows Stages 0 to 3 of the Sierpinski 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.

Figure : Stages 0, 1, 2 and 3 of the Sierpiński square carpet

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 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 red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?

Ans: (i) Number of red squares in Stages 0 to 3: Stage 0 = 1

Stage 1 = 8

Stage 2 = 8 × 8 = 64

Stage 3 = 8 × 8 × 8 = 512

So, the sequence is: 1, 8, 64, 512

(ii) Stage 4 and Stage 5: Stage 4 = 8⁴ = 4096

Stage 5 = 8⁵ = 32768

(iii) Rule (explicit and recursive): This is a geometric progression with common ratio 8.

Explicit formula:

t_n = 8ⁿ (if Stage 0 is n = 0)

So:

Stage n → t_n = 8ⁿ

Recursive formula:

t₀ = 1

t_n = 8 × t_n−1, for n ≥ 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:

s₀ = 1

s_n = (8/9) × s_n−1, for n ≥ 1

END-OF-THE-CHAPTER EXERCISES

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

Ans: We are given an arithmetic progression (AP) where:

The 11^th term t₁₁ = 38,

The 16^th term t₁₆ = 73.

We need to find the 31 st term.

Let a be the first term and d be the common difference.

t₁₁ = a + (11 – 1)d

⇒ 38 = a + 10d …(i)

And = a + (16 – 1)d …….(ii)

⇒ 73 = a+ 15d …(iii)

Subtract equation (i) from equation (ii):

(a + 15d) – (a + 10d) = 73 – 38

⇒ 5d = 35

⇒ d = 7

Substitute d = 7 into the equation (i)

t_n = a + 10d = 38

⇒ a + 10 × 7= 38

⇒ a + 70 = 38

⇒ a = 38 -70 = -32

Thus, the first term a = -32 and the common difference d = 7.

t₃₁ = a + (31 – 1) × d

= -32 + 30 × 7

= -32 + 210

= 178.

Thus, the 31st term is 178.

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

Ans: 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, …

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.)

Ans: The smallest three-digit number divisible by 7 is 105 and the largest three-digit number divisible by 7 is 994.

The terms form an arithmetic progression:

Here, first term (a) = 105,

common difference (d) = l

and last term (tn) = 994.

n_th term of an AP: tn = a + (n – 1)d

Substitute the value of a, d and tn:

⇒ 994 = 105 + (n – 1) × 7

⇒ 994 – 105 = (n – 1) × 7

⇒ n – 1 = 127

⇒ n = 128

Thus, there are 128 three-digit numbers divisible by 7.

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.)

Ans: We need multiples of 4 between 10 and 250.

First multiple of 4 greater than 10:

12 = 3 × 4

So, first term a = 12

Last multiple of 4 less than 250:

248 = 62 × 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.

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.

Ans: Let the first term of GP is a and common ratio is r.

∴ t_n = a, t₂ = ar, t₃ = ar², and t₅ = ar⁴

Given: a + ar = -A …(i)

And ar⁴ = 4 × ar² …(ii)

⇒ r² = 4

⇒ r = -2 or 2

Substitute r = 2 in equation (;):

a + a × 2 = -4

⇒ a(1 + 2) = -4

⇒ 3a = -4

⇒ a = -43

Thus, the GP is, -43, -83, -163………….

Now, substitute r = -2 in equation (i):

a + a × (-2) = -4

⇒ a – 2a = -4

⇒ -a = -4

⇒ a = 4

Thus, the GP is, 4, -8, 16,….

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

Ans: Let the number 100 be expressed as sum of n consecutive natural numbers starting from a:

a + (a + 1) + (a + 2)+…+n terms

Sum = n/2[2a + (n – 1)] = 100

So,

n(2a + n – 1) = 200

2an = 200 – n(n – 1)

Now we try values of n such that a is a natural number.

n = 1:

a = 100

So, 100

n = 5

5a + 10 = 100

5a = 90 → a = 18

So, 18 + 19 + 20 + 21 + 22

n = 8:

8a + 28 = 100

8a = 72 → a = 9

So, 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

For other values of n, a is not a natural number.

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^nd hour, 4^th hour and n^th hour?

Ans: This is a geometric progression (GP) with the first term

(a) = 30 and the common ratio (r) = 2.

The number of bacteria after n hours is: a × rⁿ

The number of bacteria after 2 hours:

30 × 22 = 30 × 4= 120.

The number of bacteria after 4 hours:

30 × 24= 30 × 16 = 480.

Therefore, the number of bacteria after n hours: 30 × 2ⁿ.

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

Ans: 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 × 5 = 12

a + 25 = 12

a = -13

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

Ans: The sum of first n natural number: Sn = n^n + ^ ^

Sum of first n natural number greater than 1000.

So, S_n > 1000

⇒ 1/2 > 1000

⇒ n(n + 1) > 2000

Since n(n + 1) is very close to n² for larger value of n, we can estimate by finding the square root of 2000.

We know that: 40² = 1600, 45² = 2025, the value n must be very close to 45.

Now we test the integers around our estimate of 45:

If n = 44: S₄₄ = 1/2 = 990

If n = 45: S₄₅ = 1/2 = 1035

This is the first sum that exceeds 1000.

So, the smallest value of n is 45.

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.

Ans: 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⋅r^n-1

So,

a_n = 2⋅4^n-1

Find which term is 131072

2⋅4^n-1 = 131072

Divide by 2:

4^n-1 = 65536

Now,

65536 = 4⁸( since 4⁸ = (2²)⁸ = 2¹⁶ = 65536)

So,

n – 1 = 8

n = 9

Recursive formula

First term: a₁ = 2

Recurrence relation: a_n = 4⋅a_n-1, n ≥ 2

11. The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.

Ans: Let the first term of GP be a and common ratio be r.

So, first three terms of GP are: a, ar, ar²

We are given that: a + ar + ar² = 13/12

⇒ a(1 + r + r²) = 13/12 ….. (i)

And a.ar.ar² = -1

⇒ (ar)³ = -1

⇒ a = -1/r …..(ii)

Substitute the value of a in equation (i):

-1/r(1 + r + r²) = 13/12

⇒ 1 + r + r² = -13r/12

⇒ 12 + 12r + 12r² = -13r

⇒ 12r² + 25r + 12 = 0

⇒ (3r + 4)(4r + 3) = 0

⇒ r = -4/3 or r = -3/4 

Case 1:

When r = -4/3 in equation (ii)

a = -1/(-43)

⇒ a = 3/4

Therefore, the first three term if the GP are 3/4, -1, 4/3

Case 2:

When r = -3/4

Substitute r = -3/4 in equation (ii)

The first three term of the GP are: 4/3, -1, 3/4

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

Ans: Let the GP be:

a, ar, ar², ar³,…

Given:

4th term = x = ar³

10th term = y = ar⁹

16th term = z = ar¹⁵

Now consider y²:

y² = (ar⁹)² = a²r¹⁸

Now consider xz:

xz = (ar³)(ar¹⁵) = a²r¹⁸

Compare:

y² = a²r¹⁸ = xz

So,

y² = xz

Conclusion:

Since y² = xz, the numbers x, y, z are in Geometric Progression (GP).

Hence proved.

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.

Ans: Let a be the first term and r be the common ratio of GP. First three term of GP be: a, ar, ar²

Given: Sum of first three terms is 26 and sum of the squares of first three terms is 364.

∴ a + ar + ar² = 26 …(i)

⇒ a(1 + r + r²) = 26

And a² + a²r² + a²r⁴ = 364 …(ii)

⇒ a²( 1 + r² + r⁴) = 364

Square both sides of equation (i)

(a + ar + ar²)² = 26²

⇒ (a² + a²r² + a²r⁴) + 2 a²(r + r² + r³) = 676

⇒ 364 + 2a²r(1 +r + r²) = 676

⇒ 2(ar)(a)(1 + r + r²) = 676 – 364 = 312

⇒ 2(ar) × 26 = 312 [∵ a(1 + r + r²) = 26]

ar = 6

So, the middle term is 6.

Since ar = 6

⇒ a = 6/r

Substitute the value of a in (i)

6/r + 6 + 6r = 26

⇒ 6/r + 6r = 26 – 6 = 20

⇒ 3/r + 3r = 10

⇒ 3 + 3r² =10r

⇒ 3r² – 10n + 3 = 0

⇒ (3r – 1) (r – 3) = 0

⇒ r = 3 or r = 1/3

If r = 3, the terms are: 6 ÷ 4, 6, 6 × 3 ⇒ 2, 6, 18

If r = 13, the terms are: 6 ÷ 13, 6, 6 × 13 ⇒ 18, 6, 2

Thus, the terms of the GP are 2, 6, and 18.

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 P_n? Can you give an explicit formula?

Ans: Given:

P₁ = 1, P₂ = 2

and for n > 2,

P_n = P₁ + P₂+⋯+P_n-1 + 1

Step 1: Find first 8 terms

We calculate step by step:

P₁ = 1

P₂ = 2

Now,

P₃ = 1 + 2 + 1 = 4

P₄ = 1 + 2 + 4 + 1 = 8

P₅ = 1 + 2 + 4 + 8 + 1=16

P₆ = 32

P₇ = 64

P₈ = 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 = 2P_n-1, n ≥ 2

with P₁ = 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:

P₁ = 1, P_n = 2 P_n−1

Explicit formula:

P_n = 2ⁿ⁻¹

15. Suppose W₁ = 1 , W₂ = 2 and for n > 2, W_n = W₁ + W₂ + … + W_n-2 + 2. Find the values of W₁, W₂, …, W₈. Do you recognise this sequence?

Ans: Given: W_n = W₁ + W₂ + … + W_n-2 + 2

Using this formula, calculate the values for Wj through W8:

W₁ = 1 [Given]

W₂ = 2 [Given]

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 = 13

W₇ = W₁ + W₂ + W₃ + W₄ + W₅ + 2 = 1 + 2 + 3 + 5 + 8 + 2 = 21

W₈ = W₁ + W₂ + W₃ + W₄ + W₅ + W₆ + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34

Thus, the values of W, and W₈ are: 1, 2, 3, 5, 8, 13, 21, 34

The sequence is a variation of the Fibonacci sequence with starting from the second 1;

i.e., the sequence is of type 1, 1, 2, 3, 5, 8, 13, 21, 34,…

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