**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. a****1**** = a****2**** = 2, a****n**** = a****n – 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, a****2****, – a****3**** ………… (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:

- a = 20 / (29 + √641), r = (29 + √641)/20
- 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 P****2**** = (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**.