Q1. End-of-Chapter Exercise 1 (Page 21): Find the 31^st term of an AP whose 11^th term is 38 and 16^th term is 73.

Answer:

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

We are given: t₁₁ = a + 10d = 38 ... (Equation 1)

t₁₆ = a + 15d = 73 ... (Equation 2)

Subtract Equation 1 from Equation 2:

(a + 15d) - (a + 10d) = 73 - 38 → 5d = 35 → d = 7.

Substitute d = 7 in Equation 1: a + 10(7) = 38 → a + 70 = 38 → a = -32.

The 31^st term is: t₃₁ = a + 30d = -32 + 30(7) = -32 + 210 = 178.

Thus, the 31^st term is 178.

Q2. End-of-Chapter Exercise 2 (Page 21): Determine the AP whose third term is 16 and whose 7^th term exceeds the 5^th term by 12.

Answer:

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

We are given: t₃ = a + 2d = 16 ... (Equation 1)

Also, t₇ - t₅ = 12.

Using the formula, t₇ = a + 6d and t₅ = a + 4d.

t₇ - t₅ = (a + 6d) - (a + 4d) = 2d = 12 → d = 6.

Substitute d = 6 in Equation 1: a + 2(6) = 16 → a + 12 = 16 → a = 4.

Thus, the AP is: 4, 10, 16, 22, 28, ...

Q3. End-of-Chapter Exercise 3 (Page 22): 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.)

Answer:

The smallest 3-digit number divisible by 7 is 105 (since 7 × 15 = 105).

The largest 3-digit number divisible by 7 is 994 (since 7 × 142 = 994).

These numbers form an AP: 105, 112, 119, ..., 994 with first term a = 105, common difference d = 7.

Let n be the number of terms. t_n = a + (n-1)d = 994.

105 + (n-1)7 = 994 → (n-1)7 = 889 → n-1 = 127 → n = 128.

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

Q4. End-of-Chapter Exercise 4 (Page 22): 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.)

Answer:

The smallest multiple of 4 greater than 10 is 12.

The largest multiple of 4 less than 250 is 248.

These numbers form an AP: 12, 16, 20, ..., 248 with first term a = 12, common difference d = 4.

Let n be the number of terms. t_n = a + (n-1)d = 248.

12 + (n-1)4 = 248 → (n-1)4 = 236 → n-1 = 59 → n = 60.

Thus, there are 60 multiples of 4 between 10 and 250.

Q5. End-of-Chapter Exercise 5 (Page 22): Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Answer:

Let a be the first term and r the common ratio.

Sum of the first two terms: a + ar = a(1+r) = -4 ... (Equation 1)

Fifth term is 4 times the third term: ar⁴ = 4(ar²) → ar⁴ = 4ar² → r² = 4 (assuming a and r are non-zero).

So, r = 2 or r = -2.

Case 1: r = 2

Substitute in Equation 1: a(1+2) = -4 → 3a = -4 → a = -4/3.

GP: -4/3, -8/3, -16/3, -32/3, -64/3, ...

Case 2: r = -2

Substitute in Equation 1: a(1-2) = -4 → -a = -4 → a = 4.

GP: 4, -8, 16, -32, 64, ...

Thus, the two possible GPs are (-4/3, -8/3, -16/3, ...) and (4, -8, 16, -32, ...).

Q6. End-of-Chapter Exercise 6 (Page 22): Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Answer:

We need to find all sequences of consecutive natural numbers that sum to 100.

Let the numbers be a, a+1, a+2, ..., a+(k-1), where a is the first term and k is the number of terms.

The sum is S = k/2 × [2a + (k-1)] = 100 → k[2a + k - 1] = 200.

We need to find integer solutions for k (number of terms) and a (first term), with a ≥ 1.

We'll find factors of 200. For each factor k, we need 2a + k - 1 = 200/k to be an even integer (since 2a is even).

• k = 1: 2a = 200 → a = 100. Sequence: 100
• k = 2: 2a + 1 = 100 → 2a = 99 (not integer)
• k = 4: 2a + 3 = 50 → 2a = 47 (not integer)
• k = 5: 2a + 4 = 40 → 2a = 36 → a = 18. Sequence: 18, 19, 20, 21, 22 (sum = 100)
• k = 8: 2a + 7 = 25 → 2a = 18 → a = 9. Sequence: 9, 10, 11, 12, 13, 14, 15, 16 (sum = 100)
• k = 10: 2a + 9 = 20 → 2a = 11 (not integer)
• k = 16: 2a + 15 = 12.5 → 2a = -2.5 (not possible, a negative)

Thus, the possible ways are:

• 100
• 18 + 19 + 20 + 21 + 22
• 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Q7. End-of-Chapter Exercise 7 (Page 22): 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?

Answer:

The number of bacteria forms a GP with first term a = 30 and common ratio r = 2 (since it doubles every hour).

The number of bacteria at the end of the n^th hour is given by:

t_n = 30 × 2ⁿ

• End of 2^nd hour: t₂ = 30 × 2² = 30 × 4 = 120 bacteria
• End of 4^th hour: t₄ = 30 × 2⁴ = 30 × 16 = 480 bacteria
• End of n^th hour: t_n = 30 × 2ⁿ bacteria

Q8. End-of-Chapter Exercise 8 (Page 22): 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.

Answer:

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

Given: t₄ + t₈ = 24 → (a+3d) + (a+7d) = 2a + 10d = 24 ... (Equation 1)

Also: t₆ + t₁₀ = 44 → (a+5d) + (a+9d) = 2a + 14d = 44 ... (Equation 2)

Subtract Equation 1 from Equation 2:

(2a + 14d) - (2a + 10d) = 44 - 24 → 4d = 20 → d = 5.

Substitute d = 5 in Equation 1: 2a + 10(5) = 24 → 2a + 50 = 24 → 2a = -26 → a = -13.

Thus, the first three terms are:

• t₁ = a = -13
• t₂ = a + d = -13 + 5 = -8
• t₃ = a + 2d = -13 + 10 = -3

The AP is: -13, -8, -3, ...

Q9. End-of-Chapter Exercise 9 (Page 22): Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Answer:

The sum of the first n natural numbers is S_n = n(n+1)/2.

We need the smallest n such that S_n > 1000.

n(n+1)/2 > 1000 → n(n+1) > 2000.

Let's find n such that n² + n - 2000 > 0.

√2000 ≈ 44.72, so n is around 44.

Check n = 44: 44(45)/2 = 990.

Check n = 45: 45(46)/2 = 1035.

Since 990 < 1000 and 1035 > 1000, the smallest n is 45.

Q10. End-of-Chapter Exercise 10 (Page 22): Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the n^th term.

Answer:

The GP is 2, 8, 32, ... with first term a = 2 and common ratio r = 8/2 = 4.

Explicit formula: t_n = 2 × 4^n-1

Recursive formula: t₁ = 2, t_n = 4 × t_n-1 for n ≥ 2.

Which term is 131072?

Set t_n = 2 × 4^n-1 = 131072 → 4^n-1 = 65536.

We know 4⁸ = 65536 (since 4⁸ = 2¹⁶ = 65536). So, n-1 = 8 → n = 9.

Thus, 131072 is the 9th term of the GP.

Q11. End-of-Chapter Exercise 11 (Page 22): 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.

Answer:

Let the first three terms of the GP be a, ar, ar².

Given: a + ar + ar² = a(1 + r + r²) = 13/12 ... (Equation 1)

Also, a × ar × ar² = a³r³ = (ar)³ = -1 → ar = -1 ... (Equation 2)

From Equation 2, a = -1/r.

Substitute in Equation 1: (-1/r)(1 + r + r²) = 13/12 → -(1 + r + r²)/r = 13/12.

Multiply both sides by -12r: 12(1 + r + r²) = -13r → 12 + 12r + 12r² + 13r = 0 → 12r² + 25r + 12 = 0.

Factor the quadratic: (3r + 4)(4r + 3) = 0 → r = -4/3 or r = -3/4.

Case 1: r = -4/3

a = -1/r = -1/(-4/3) = 3/4.

Terms: 3/4, (3/4)(-4/3) = -1, (-1)(-4/3) = 4/3.

Case 2: r = -3/4

a = -1/r = -1/(-3/4) = 4/3.

Terms: 4/3, (4/3)(-3/4) = -1, (-1)(-3/4) = 3/4.

Thus, the GP is either (3/4, -1, 4/3, ...) or (4/3, -1, 3/4, ...).

Q12. End-of-Chapter Exercise 12 (Page 22): 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.

Answer:

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

We are given:

• t₄ = ar³ = x
• t₁₀ = ar⁹ = y
• t₁₆ = ar¹⁵ = z

To prove that x, y, z are in GP, we need to show y² = xz.

y² = (ar⁹)² = a²r¹⁸

xz = (ar³)(ar¹⁵) = a²r¹⁸

Since y² = xz, the numbers x, y, z are indeed in Geometric Progression.

Q13. End-of-Chapter Exercise 13 (Page 22): 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.

Answer:

Let the first three terms be a, ar, ar².

Given: a + ar + ar² = a(1 + r + r²) = 26 ... (Equation 1)

And a² + a²r² + a²r⁴ = a²(1 + r² + r⁴) = 364 ... (Equation 2)

From Equation 1, a = 26 / (1 + r + r²).

Substitute in Equation 2: [26² / (1 + r + r²)²] × (1 + r² + r⁴) = 364.

676(1 + r² + r⁴) = 364(1 + r + r²)².

Divide both sides by 4: 169(1 + r² + r⁴) = 91(1 + r + r²)².

Expanding: 169 + 169r² + 169r⁴ = 91(1 + r² + r⁴ + 2r + 2r² + 2r³) = 91(1 + 2r + 3r² + 2r³ + r⁴).

169 + 169r² + 169r⁴ = 91 + 182r + 273r² + 182r³ + 91r⁴.

Bringing all to one side: 78 - 182r - 104r² - 182r³ + 78r⁴ = 0 → 78r⁴ - 182r³ - 104r² - 182r + 78 = 0.

Divide by 2: 39r⁴ - 91r³ - 52r² - 91r + 39 = 0.

We can find r by trial. Try r = 3: 39(81) - 91(27) - 52(9) - 91(3) + 39 = 3159 - 2457 - 468 - 273 + 39 = 0. So r = 3 is a root.

Try r = 1/3: 39(1/81) - 91(1/27) - 52(1/9) - 91(1/3) + 39 = 13/27 - 91/27 - 52/9 - 91/3 + 39 = (13 - 91)/27 - 52/9 - 273/9 + 39 = -78/27 - 325/9 + 39 = -26/9 - 325/9 + 351/9 = 0. So r = 1/3 is also a root.

If r = 3, from Equation 1: a(1 + 3 + 9) = 26 → 13a = 26 → a = 2. Terms: 2, 6, 18.

If r = 1/3, from Equation 1: a(1 + 1/3 + 1/9) = 26 → a(13/9) = 26 → a = 18. Terms: 18, 6, 2.

Thus, the terms are 2, 6, 18 or 18, 6, 2.

Q14. End-of-Chapter Exercise 14 (Page 22): 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?

Answer:

Finding the first 8 terms:

• P₁ = 1
• P₂ = 2
• P₃ = P₁ + P₂ + 1 = 1 + 2 + 1 = 4
• P₄ = P₁ + P₂ + P₃ + 1 = 1 + 2 + 4 + 1 = 8
• P₅ = 1 + 2 + 4 + 8 + 1 = 16
• P₆ = 1 + 2 + 4 + 8 + 16 + 1 = 32
• P₇ = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64
• P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128

Thus, P₁, P₂, ..., P₈ are: 1, 2, 4, 8, 16, 32, 64, 128.

Simpler recursive formula:

For n ≥ 2, P_n = 2 × P_n-1. Since each term is double the previous one.

Explicit formula:

The sequence is a GP with a = 1 and r = 2. For n ≥ 1, P_n = 2^n-1.

Q15. End-of-Chapter Exercise 15 (Page 22): 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?

Answer:

Finding the first 8 terms:

• W₁ = 1
• W₂ = 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 = 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, W₁, W₂, ..., W₈ are: 1, 2, 3, 5, 8, 13, 21, 34.

Recognition: This is the Virahanka-Fibonacci sequence (with a slight variation; here it starts 1, 2, 3, 5, ... but the standard Fibonacci is 1, 1, 2, 3, 5, ...). The sequence follows the pattern where each term is the sum of the two preceding terms.

We can see that for n ≥ 3, W_n = W_n-1 + W_n-2.