Sequences and Series

Chapter 8.1: Sequences

• Sequence: An ordered collection of numbers or terms.
• Arithmetic Sequence: A sequence where the difference between consecutive terms is constant (common difference, d).
• Geometric Sequence: A sequence where the ratio between consecutive terms is constant (common ratio, r).
• General Term (nth Term): The formula for finding any term in a sequence.

Chapter 8.2: Series

• Series: 
• Arithmetic Series: The sum of terms in an arithmetic sequence.
• Geometric Series: 
• Sum of an Arithmetic Series: Sn = n/2 * [2a + (n-1)d]
• Sum of a Geometric Series: Sn = a(r^n – 1) / (r – 1) (when r ≠ 1)

Chapter 8.3: Arithmetic and Geometric Progressions

• Properties and Applications: Discusses various properties and applications of arithmetic and geometric progressions, such as finding specific terms, sums, and solving problems involving these sequences.

Key Concepts:

• Sequences and series
• Arithmetic and geometric sequences
• General terms, common difference, and common ratio
• Sum formulas for arithmetic and geometric series
• Applications of sequences and series in real-world problems

Exercise 8.1

1.Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: 

1. an = n (n + 2)

Ans : 

1. For n = 1: a1 = 1(1 + 2) = 1 * 3 = 3

2. For n = 2: a2 = 2(2 + 2) = 2 * 4 = 8

3. For n = 3: a3 = 3(3 + 2) = 3 * 5 = 15

4. For n = 4: a4 = 4(4 + 2) = 4 * 6 = 24

5. For n = 5: a5 = 5(5 + 2) = 5 * 7 = 35

2.  an = n/n+1

Ans : 

1. For n = 1: a1 = 1 / (1 + 1) = 1 / 2

2. For n = 2: a2 = 2 / (2 + 1) = 2 / 3

3. For n = 3: a3 = 3 / (3 + 1) = 3 / 4

4. For n = 4: a4 = 4 / (4 + 1) = 4 / 5

5. For n = 5: a5 = 5 / (5 + 1) = 5 / 6

3. an = 2 n 

Ans : 

1. For n = 1: a1 = 2^1 = 2

2. For n = 2: a2 = 2^2 = 4

3. For n = 3: a3 = 2^3 = 8

4. For n = 4: a4 = 2^4 = 16

5. For n = 5: a5 = 2^5 = 32

4.  an = 2n−3/6

Ans : 

1. For n = 1: a1 = (2 * 1 – 3) / 6 = -1 / 6

2. For n = 2: a2 = (2 * 2 – 3) / 6 = 1 / 6

3. For n = 3: a3 = (2 * 3 – 3) / 6 = 3 / 6 = 1/2

4. For n = 4: a4 = (2 * 4 – 3) / 6 = 5 / 6

5. For n = 5: a5 = (2 * 5 – 3) / 6 = 7 / 6

5. 

Ans : 

1. For n = 1: a1 = (-1)^(1-1) * (5 * 1 + 1) = 1 * 6 = 6

2. For n = 2: a2 = (-1)^(2-1) * (5 * 2 + 1) = -1 * 11 = -11

3. For n = 3: a3 = (-1)^(3-1) * (5 * 3 + 1) = 1 * 16 = 16

4. For n = 4: a4 = (-1)^(4-1) * (5 * 4 + 1) = -1 * 21 = -21

5. For n = 5: a5 = (-1)^(5-1) * (5 * 5 + 1) = 1 * 26 = 26

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n th terms are: 

7. an = 4n – 3; a17, a24

Ans : 

For n = 17:

a17 = 4(17) – 3 = 68 – 3 = 65

For n = 24:

a24 = 4(24) – 3 = 96 – 3 = 93

8. 

Ans : 

To find the 7th term of the sequence, we need to substitute n = 7 into the given formula for the nth term:

an = n² / (2n)

Substituting n = 7:

a7 = 7² / (2 * 7)

a7 = 49 / 14

a7 = 7/2

Therefore, the 7th term of the sequence is 7/2.

9. an = (–1)n – 1n 3 ; a9

Ans : 

To find the 9th term of the sequence defined by an = (-1)^(n-1) * n^3, we simply substitute n = 9 into the formula:

a9 = (-1)^(9-1) * 9^3

a9 = (-1)^8 * 729

a9 = 1 * 729

a9 = 729

Therefore, the 9th term of the sequence is 729.

10. 

Ans : 

To find the 20th term of the sequence, we need to substitute n = 20 into the given formula for the nth term:

an = n(n-2) / (n+3)

Substituting n = 20:

a20 = 20(20-2) / (20+3)

a20 = 20(18) / 23

a20 = 360 / 23

Therefore, the 20th term of the sequence is 360/23.

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 

11. a1 = 3, an = 3an – 1 + 2 for all n > 1

Ans : 

1. First term (a1) is given:

a1 = 3

2. Second term (a2):

a2 = 3a1 + 2 = 3 * 3 + 2 = 9 + 2 = 11

3. Third term (a3):

a3 = 3a2 + 2 = 3 * 11 + 2 = 33 + 2 = 35

4. Fourth term (a4):

a4 = 3a3 + 2 = 3 * 35 + 2 = 105 + 2 = 107

5. Fifth term (a5):

a5 = 3a4 + 2 = 3 * 107 + 2 = 321 + 2 = 323

12. 

Ans : 

1. First term (a1) is given:

a1 = -1

2. Second term (a2):

a2 = (a1) / 2 = -1 / 2

3. Third term (a3):

a3 = (a2) / 3 = (-1/2) / 3 = -1/6

4. Fourth term (a4):

a4 = (a3) / 4 = (-1/6) / 4 = -1/24

5. Fifth term (a5):

a5 = (a4) / 5 = (-1/24) / 5 = -1/120

13. a1 = a2 = 2, an = an – 1 – 1, n > 2.

Ans : 

1. First term (a1) is given:

a1 = 2

2. Second term (a2) is given:

a2 = 2

3. Third term (a3):

a3 = a2 – 1 = 2 – 1 = 1

4. Fourth term (a4):

a4 = a3 – 1 = 1 – 1 = 0

5. Fifth term (a5):

a5 = a4 – 1 = 0 – 1 = -1

14. The Fibonacci sequence is defined by 1 = a1 = a2 and an = an – 1 + an – 2 n > 2.  Find an+1/an, for n = 1, 2, 3, 4, 5.

Ans : 

Given:

a1 = 1 a2 = 1 an = an-1 + an-2 for n > 2

Calculating the first few terms:

a3 = a2 + a1 = 1 + 1 = 2 a4 = a3 + a2 = 2 + 1 = 3 a5 = a4 + a3 = 3 + 2 = 5 a6 = a5 + a4 = 5 + 3 = 8

Now, we can calculate an+1/an for n = 1, 2, 3, 4, 5:

• For n = 1: a2/a1 = 1/1 = 1
• For n = 2: a3/a2 = 2/1 = 2
• For n = 3: a4/a3 = 3/2
• For n = 4: a5/a4 = 5/3
• For n = 5: a6/a5 = 8/5

Exercise 8.2

1. 

Ans  : 

.We are given a geometric progression (GP) with the first few terms as:

5/2, 5/4, 5/8, …

Identifying the common ratio (r):

The common ratio (r) is the ratio between any two consecutive terms. In this case:

r = (5/4) / (5/2) = (5/4) * (2/5) = 1/2

General term (nth term) of a GP:

an = a1 * r^(n-1)

Where:

• a1 is the first term
• r is the common ratio
• n is the term number

Given:

• a1 = 5/2
• r = 1/2

Finding the 20th term:

a20 = (5/2) * (1/2)^(20-1) a20 = (5/2) * (1/2)^19 a20 = 5 * 2^(-20)

Finding the nth term:

an = (5/2) * (1/2)^(n-1)

Therefore, the 20th term of the GP is 5 * 2^(-20), and the nth term is (5/2) * (1/2)^(n-1).

2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Ans : 

an = a1 * r^(n-1)

Where:

• a1 is the first term
• r is the common ratio
• n is the term number

We are given:

• a8 = 192
• r = 2

a12 = a1 * r^(12-1)

We can find a1 by using the given information about a8:

a8 = a1 * r^(8-1) 192 = a1 * 2^7 a1 = 192 / 2^7 a1 = 3

Now, we can substitute a1 and r into the formula for a12:

a12 = 3 * 2^(12-1) a12 = 3 * 2^11 a12 = 6144

Therefore, the 12th term of the GP is 6144.

3. The 5th, 8th and 11 th terms of a G.P. are p, q and s, respectively. Show that q 2 = ps.

Ans : 

Let the first term of the GP be a and the common ratio be r. Then, we have:

• a5 = p
• a8 = q
• a11 = s

Using the formula for the nth term of a GP:

an = a1 * r^(n-1)

We can write:

• p = a1 * r^4
• q = a1 * r^7
• s = a1 * r^10

Now, we need to show that q^2 = ps.

Substituting the values of p, q, and s:

(a1 * r^7)^2 = (a1 * r^4) * (a1 * r^10)

Simplifying:

a1^2 * r^14 = a1^2 * r^14

Therefore, q^2 = ps is proven.

4. that q 2 = ps. 4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term. 

Ans : 

Then, we have:

• a1 = -3 (given)
• a4 = a1 * r^3 (formula for nth term of a GP)

We are also given that a4 is the square of a2:

a4 = (a2)^2

Substituting the values of a1 and a4:

-3 * r^3 = (-3 * r)^2

Simplifying:

-3 * r^3 = 9 * r^2

Dividing both sides by -3 * r^2:

r = -3

Now, we can find the 7th term using the formula for the nth term:

a7 = a1 * r^(7-1)

a7 = -3 * (-3)^6

a7 = -3 * 729

a7 = -2187

Therefore, the 7th term of the GP is -2187.

5. Which term of the following sequences :

(a) 2, 2√2, 4, ……….. is 128 ?

(b) √3, 3, 3√3, ………… is 729?

(c) 1/3,1/9,1/27,… is 1/19683?

Ans : 

(a) Finding the term with value 128

Let the nth term be 128. Then:

128 = 2 * (√2)^(n-1)

Solving for n:

(√2)^(n-1) = 64 

(√2)^(n-1) = (√2)^12

 n – 1 = 12 

n = 13

Therefore, the 13th term of sequence (a) is 128.

(b) Finding the term with value 729

Let the nth term be 729. Then:

729 = √3 * (3)^(n-1)

Solving for n:

(3)^(n-1) = 729 / √3 

(3)^(n-1) = (3)^(3/2) * (3)^(3/2) 

(3)^(n-1) = (3)^(3/2 + 3/2) 

(3)^(n-1) = (3)^3 

n – 1 = 3 

n = 4

Therefore, the 4th term of sequence (b) is 729.

(c) Finding the term with value 1/19683

Let the nth term be 1/19683. Then:

1/19683 = (1/3) * (1/3)^(n-1)

Solving for n:

(1/3)^(n-1) = 1/19683 * 3 

(1/3)^(n-1) = 1/6561 

(1/3)^(n-1) = (1/3)^8 

n – 1 = 8 

n = 9

Therefore, the 9th term of sequence (c) is 1/19683.

6. For what values of x, the numbers – 2/7, x, – 7/2 are in GP.?

Ans : 

To determine if the numbers -2/7, x, and -7/2 are in geometric progression (GP), we need to check if the common ratio (r) is the same between any two consecutive terms.

The common ratio (r) is calculated as:

r = (second term) / (first term)

In this case:

r = x / (-2/7) = -7x/2

Now, we need to check if this common ratio is consistent between the second and third terms:

r = (-7/2) / x = -7 / (2x)

Since the common ratio must be the same for all consecutive terms, we can set the two expressions for r equal to each other:

-7x/2 = -7 / (2x)

Multiplying both sides by 2x:

-7x² = -7

Dividing both sides by -7:

x² = 1

Taking the square root of both sides:

x = ±1

Therefore, the numbers -2/7, x, and -7/2 are in GP if x is either 1 or -1.

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 

7. 0.15, 0.015, 0.0015, … 20 terms

Ans : 

Sn = a(1 – r^n) / (1 – r)

where:

• Sn is the sum of the first n terms
• a is the first term
• r is the common ratio  
• n is the number of terms

In this case:

• a = 0.15
• r = 0.015 / 0.15 = 0.1
• n = 20

Substituting these values into the formula:

S20 = 0.15(1 – 0.1^20) / (1 – 0.1)

S20 = 0.15(1 – 0.0000000001) / 0.9

S20 = 0.15 * 0.9999999999 / 0.9

S20 = 0.1666666666

Therefore, the sum of the first 20 terms of the given geometric progression is approximately 0.1666666666.

8. 

Ans : 

The given G.P. is √7, √21, 3√7 ……………

Here, a = √7

9.  1, – a, a2, – a3 ………… (if a ≠ 1)

Ans : 

10. x3, x5, x7 … (if x ≠ ± 1).

Ans : 

Sn = a(1 – r^n) / (1 – r)

In this case:

• a = x^3
• r = x^2
• n is the number of terms we want to sum

Substituting these values into the formula:

Sn = x^3(1 – (x^2)^n) / (1 – x^2)

Sn = x^3(1 – x^(2n)) / (1 – x^2)

Therefore, the sum to n terms of the given geometric progression is:

Sn = x^3(1 – x^(2n)) / (1 – x^2)

11. Evaluate 

∑−k=111 (2 + 3k)

Ans : 

12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Ans : 

• a1 = a
• a2 = ar
• a3 = ar^2

We are given that the sum of the first three terms is 39/10:

a + ar + ar^2 = 39/10

We are also given that the product of the first three terms is 1:

a * ar * ar^2 = 1

Simplifying the second equation:

a^3 * r^3 = 1

Taking the cube root of both sides:

a * r = 1

Substituting a * r = 1 into the first equation:

a + 1 + 1/r = 39/10

Multiplying both sides by 10r:

10ar + 10r + 10 = 39r

Rearranging:

10ar – 29r + 10 = 0

We can solve this quadratic equation for r using the quadratic formula:

r = (-b ± √(b^2 – 4ac)) / 2a

where a = 10, b = -29, and c = 10.

Substituting these values into the formula:

r = (29 ± √(29^2 – 4 * 10 * 10)) / 2 * 10

r = (29 ± √641) / 20

Therefore, the common ratio can be either (29 + √641)/20 or (29 – √641)/20.

To find the first term (a), we can substitute the common ratio into the equation a * r = 1:

a = 1 / r

Substituting the two possible values of r:

a = 1 / ((29 + √641)/20) = 20 / (29 + √641)

or

a = 1 / ((29 – √641)/20) = 20 / (29 – √641)

Therefore, there are two possible geometric progressions:

1. a = 20 / (29 + √641), r = (29 + √641)/20
2. a = 20 / (29 – √641), r = (29 – √641)/20

You can verify that both of these progressions satisfy the given condition

13. How many terms of G.P. 3, 32 , 33 , … are needed to give the sum 120?

Ans : 

Sn = a(1 – r^n) / (1 – r)

where:

• Sn is the sum of the first n terms
• a is the first term
• r is the common ratio  
• n is the number of terms

In this case:

• a = 3
• r = 3
• Sn = 120

Substituting these values into the formula:

120 = 3(1 – 3^n) / (1 – 3)

Simplifying:

120 = 3(1 – 3^n) / (-2)

Multiplying both sides by -2:

-240 = 3(1 – 3^n)

Dividing both sides by 3:

-80 = 1 – 3^n

Adding 1 to both sides:

-79 = -3^n

Taking the logarithm of both sides (base 3):

log3(-79) = n

14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Ans : 

• a1 = a
• a2 = ar
• a3 = ar^2

We are given that the sum of the first three terms is 16:

a + ar + ar^2 = 16

We are also given that the sum of the next three terms is 128:

ar^3 + ar^4 + ar^5 = 128

Dividing the second equation by the first equation:

(ar^3 + ar^4 + ar^5) / (a + ar + ar^2) = 128 / 16

Simplifying:

r^3 = 8

Taking the cube root of both sides:

r = 2

Now, we can substitute r = 2 into the first equation:

a + 2a + 4a = 16

7a = 16

a = 16/7

Therefore, the first term of the GP is 16/7 and the common ratio is 2.

To find the sum to n terms of the GP, we can use the formula:

Sn = a(1 – r^n) / (1 – r)

Substituting the values of a and r:

Sn = (16/7)(1 – 2^n) / (1 – 2)

Sn = (16/7)(1 – 2^n) / (-1)

Sn = -16/7 * (1 – 2^n)

Therefore, the sum to n terms of the GP is -16/7 * (1 – 2^n).

15. Given a G.P. with a = 729 and 7th term 64, determine S7

Ans : 

an = a * r^(n-1)

Where:

• an is the nth term
• a is the first term
• r is the common ratio
• n is the term number

Finding the common ratio (r):

We know a7 = 64, so we can substitute the values into the formula:

64 = 729 * r^(7-1)

64/729 = r^6

Taking the sixth root of both sides:

(2/3)^6 = r^6

r = 2/3

Finding the sum of the first 7 terms (S7):

Sn = a(1 – r^n) / (1 – r)

Substituting the values:

S7 = 729(1 – (2/3)^7) / (1 – 2/3)

S7 = 729(1 – 128/2187) / (1/3)

S7 = 729(2059/2187) * 3

S7 = 2059

Therefore, the sum of the first 7 terms (S7) is 2059.

16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Ans : 

• a1 = a
• a2 = ar
• a3 = ar^2
• a4 = ar^3
• a5 = ar^4

We are given that the sum of the first two terms is -4:

a + ar = -4

We are also given that the fifth term is 4 times the third term:

ar^4 = 4 * ar^2

Dividing both sides by ar^2:

r^2 = 4

Taking the square root of both sides:

r = ±2

Now, we can substitute the value of r into the equation a + ar = -4:

Case 1: r = 2

a + 2a = -4

3a = -4

a = -4/3

Case 2: r = -2

a – 2a = -4

-a = -4

a = 4

17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P

Ans : 

Then, we have:

• a4 = a * r^3
• a10 = a * r^9
• a16 = a * r^15

We need to prove that x, y, and z are in GP, which means that:

y^2 = xz

Substituting the values of x, y, and z:

(a * r^9)^2 = (a * r^3) * (a * r^15)

Simplifying:

a^2 * r^18 = a^2 * r^18

Therefore, y^2 = xz is true for any GP, and thus, x, y, and z are in GP.

18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Ans : 

Let’s analyze the first few terms:

• 8 = 8 * 1
• 88 = 8 * 11
• 888 = 8 * 111
• 8888 = 8 * 1111

We can see that each term is 8 times a number that increases by 10 each time: 1, 11, 111, 1111, …

Sn = 8(1 + 11 + 111 + … + 11…1 (n digits))

To find the sum of the series 1 + 11 + 111 + …, we can use the formula for the sum of a geometric series:

Sn = a(1 – r^n) / (1 – r)

where:

• a is the first term (1 in this case)
• r is the common ratio (10 in this case)
• n is the number of terms

Substituting the values:

Sn = 1(1 – 10^n) / (1 – 10)

Sn = (1 – 10^n) / (-9)

Sn = (10^n – 1) / 9

Now, we can substitute this value back into the original expression for the sum of the given sequence:

Sn = 8(10^n – 1) / 9

Therefore, the sum to n terms of the sequence 8, 88, 888, 8888, … is 8(10^n – 1) / 9.

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 .

Ans : 

Let’s denote the two sequences as:

• Sequence 1: 2, 4, 8, 16, 32
• Sequence 2: 128, 32, 8, 2, 1/2

We need to find the sum of the products of corresponding terms of these sequences.

Let’s calculate the products of corresponding terms:

• 2 * 128 = 256
• 4 * 32 = 128
• 8 * 8 = 64
• 16 * 2 = 32
• 32 * 1/2 = 16

Now, let’s sum these products:

256 + 128 + 64 + 32 + 16 = 496

Therefore, the sum of the products of the corresponding terms of the given sequences is 496.

20. Show that the products of the corresponding terms of the sequences a, ar, ar2 , …arn – 1 and A, AR, AR2 , … ARn – 1 form a G.P, and find the common ratio.

Ans : 

• P1 = a * A
• P2 = ar * AR
• P3 = ar^2 * AR^2
• Pn = arn-1 * AR^(n-1)

To show that these products form a GP, we need to prove that the ratio between any two consecutive products is constant.

Let’s calculate the ratio between P2 and P1:

P2 / P1 = (ar * AR) / (a * A) = r * R

Similarly, we can calculate the ratio between P3 and P2:

P3 / P2 = (ar^2 * AR^2) / (ar * AR) = r * R

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Ans : 

• a1 = a
• a2 = ar
• a3 = ar^2
• a4 = ar^3

ar^2 = a + 9

ar = ar^3 + 18

Simplifying the second equation:

ar – ar^3 = 18

ar(1 – r^2) = 18

Dividing both sides by ar:

1 – r^2 = 18/ar

Substituting ar = a + 9 from the first equation:

1 – r^2 = 18 / (a + 9)

Rearranging:

r^2 + 18/(a + 9) – 1 = 0

Multiplying both sides by (a + 9):

r^2(a + 9) + 18 – (a + 9) = 0

Expanding:

ar^2 + 9r^2 + 18 – a – 9 = 0

Substituting a = ar^2 – 9 from the first equation:

(ar^2 – 9)r^2 + 9r^2 + 18 – (ar^2 – 9) – 9 = 0

Simplifying:

ar^4 – 9r^2 + 9r^2 + 18 – ar^2 + 9 – 9 = 0

ar^4 – ar^2 + 18 = 0

We can solve this quadratic equation for r^2 using the quadratic formula:

r^2 = (a ± √(a^2 – 4 * a * 18)) / (2 * a)

r^2 = (a ± √(a^2 – 72a)) / (2a)

Since r^2 must be positive, we can only consider the positive square root:

r^2 = (a + √(a^2 – 72a)) / (2a)

Now, we can substitute r^2 into the equation ar^2 = a + 9:

a * ((a + √(a^2 – 72a)) / (2a)) = a + 9

Simplifying:

(a + √(a^2 – 72a)) / 2 = a + 9

Multiplying both sides by 2:

a + √(a^2 – 72a) = 2a + 18

Subtracting a from both sides:

√(a^2 – 72a) = a + 18

Squaring both sides:

a^2 – 72a = a^2 + 36a + 324

Simplifying:

-108a = 324

a = -3

Now, we can substitute a = -3 into the equation r^2 = (a + √(a^2 – 72a)) / (2a):

r^2 = (-3 + √((-3)^2 – 72 * (-3))) / (2 * (-3))

r^2 = (-3 + √243) / (-6)

r^2 = (9 – 3√3) / 2

Taking the square root of both sides:

r = ±√((9 – 3√3) / 2)

Since r must be positive (because the terms are increasing), we take the positive square root:

r = √((9 – 3√3) / 2)

Therefore, the four numbers forming the geometric progression are:

-3, -3 * √((9 – 3√3) / 2), -3 * (9 – 3√3), -3 * (9 – 3√3) * √((9 – 3√3) / 2)

22. 

Ans : 

• ap = A * R^(p-1)
• aq = A * R^(q-1)
• ar = A * R^(r-1)

We need to prove that:

a^(q-r) * b^(r-p) * c^(p-q) = 1

Substituting the values of a, b, and c:

(A * R^(p-1))^(q-r) * (A * R^(q-1))^(r-p) * (A * R^(r-1))^(p-q) = 1

Simplifying:

A^(q-r+r-p+p-q) * R^((p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)) = 1

Since the exponents of A and R sum to 0, we have:

A^0 * R^0 = 1

Therefore, the equation is true, and we have proven that:

a^(q-r) * b^(r-p) * c^(p-q) = 1

23. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.

Ans : 

The first term of the G.P. is a and the last term is b.

Therefore, the G.P. is a, ar, ar2, ar3, …………. arn – 1, where r is the common ratio.

b = arn – 1 …………..(i)

P = Product of n terms

= (a) (ar) (ar2) … (arn – 1)

= (a × a × ……… a) (r × r2 × ……….. rn – 1)

= an r1 + 2 + ………. + (n – 1)

Here, 1, 2, ………. (n – 1) is an A.P.

24. 

Ans : 

Sum of the first n terms (Sn):

Sn = a(1 – r^n) / (1 – r)

Sum of terms from (n+1)th to (2n)th terms (S2n – Sn):

S2n – Sn = a(r^n – r^(2n)) / (1 – r)

Ratio of the two sums:

(Sn) / (S2n – Sn) = [a(1 – r^n) / (1 – r)] / [a(r^n – r^(2n)) / (1 – r)]

Simplifying:

(Sn) / (S2n – Sn) = (1 – r^n) / (r^n – r^(2n))

(Sn) / (S2n – Sn) = (1 – r^n) / (r^n(1 – r))

(Sn) / (S2n – Sn) = (1 – r^n) / (r^n) * (1 / (1 – r))

(Sn) / (S2n – Sn) = 1/r^n

Therefore, the ratio of the sum of the first n terms of a GP to the sum of terms from (n+1)th to (2n)th term is 1/r^n.

25. 

Ans : 

• b = ar
• c = ar^2
• d = ar^3

Substituting these values into the given expression:

(a^2 + (ar)^2 + (ar^2)^2)(b^2 + (ar)^2 + (ar^2)^2) = (ab + bc + cd)^2

Simplifying:

(a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6) = (a^2r + a^2r^3 + a^2r^5)^2

Factoring out a^2 from both terms:

a^2(1 + r^2 + r^4) * a^2r^2(1 + r^2 + r^4) = (a^2r(1 + r^2 + r^4))^2

Simplifying further:

a^4 * r^2 * (1 + r^2 + r^4)^2 = a^4 * r^2 * (1 + r^2 + r^4)^2

Therefore, the equation is true for any GP, and the given expression is always equal to (ab + bc + cd)^2.

26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P

Ans : 

Let the two numbers to be inserted be x and y.

Then, the sequence becomes: 3, x, y, 81

Since this is a GP, the common ratio (r) is the same between any two consecutive terms.

Therefore, we have:

r = x/3 = y/x = 81/y

x = 3r y = xr = 3r^2

Substituting these values into the third equation:

81/y = 3r

81 = 3r * y

81 = 3r * 3r^2

81 = 9r^3

Dividing both sides by 9:

9 = r^3

Taking the cube root of both sides:

r = 3

Now, we can find x and y using the values of r:

x = 3r = 3 * 3 = 9

y = 3r^2 = 3 * 3^2 = 27

Therefore, the two numbers to be inserted between 3 and 81 to form a GP are 9 and 27.

27. Find the value of n so that an+1+bn+1/ an+bn  may be the geometric mean between a and b.

Ans : 

28. 

Ans :

29. If A and G be AM. and G.M., respectively between two positive numbers, prove that the numbers are A ± (A+G)(A−G).

Ans : 

30. 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 2 nd hour, 4th hour and n th hour ? 

Ans : 

The growth of bacteria in this scenario can be modeled using a geometric progression. The initial number of bacteria is the first term (a1), and the common ratio (r) is 2 since the number doubles each hour.

At the end of the 2nd hour:

The number of bacteria will be the second term (a2) of the geometric progression.

a2 = a1 * r^(2-1) = 30 * 2^1 = 30 * 2 = 60

At the end of the 4th hour:

The number of bacteria will be the fourth term (a4) of the geometric progression.

a4 = a1 * r^(4-1) = 30 * 2^3 = 30 * 8 = 240

At the end of the nth hour:

The number of bacteria will be the nth term (an) of the geometric progression.

an = a1 * r^(n-1) = 30 * 2^(n-1)

Therefore, the number of bacteria present at the end of the 2nd hour, 4th hour, and nth hour are 120, 480, and 30 * 2^(n-1) respectively.

31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Ans : 

A = P(1 + r/n)^(nt)

Substituting the values into the formula:

A = 500(1 + 0.10/1)^(1 * 10)

A = 500(1.1)^10

Using a calculator to compute (1.1)^10, we get:

A ≈ 500 * 2.59374

A ≈ 1296.87

32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. 

Ans : 

Given:

• Arithmetic Mean (AM) = (α + β) / 2 = 8
• Geometric Mean (GM) = √(αβ) = 5

From the AM equation, we get:

α + β = 16

From the GM equation, we get:

αβ = 25

Now, we know that a quadratic equation with roots α and β can be written in the form:

(x – α)(x – β) = 0

Expanding this equation:

x^2 – (α + β)x + αβ = 0

Substituting the values of α + β and αβ:

x^2 – 16x + 25 = 0

Therefore, the quadratic equation is x^2 – 16x + 25 = 0.