Q1. Exercise Set 8.2, Question 1 (Page 185): Find the 10^th and 26^th terms of the AP: 3, 8, 13, 18, ...

Answer:

The AP is 3, 8, 13, 18, ... with first term a = 3 and common difference d = 5.

The explicit formula for the n^th term is: t_n = a + (n-1)d = 3 + (n-1)5 = 5n - 2.

• 10^th term: t₁₀ = 5(10) - 2 = 50 - 2 = 48
• 26^th term: t₂₆ = 5(26) - 2 = 130 - 2 = 128

Q2. Exercise Set 8.2, Question 2 (Page 12): Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.

Answer:

The AP is 21, 18, 15, ... with first term a = 21 and common difference d = 18 - 21 = -3.

The explicit formula is: t_n = a + (n-1)d = 21 + (n-1)(-3) = 21 - 3n + 3 = 24 - 3n.

Which term is -81?

Set 24 - 3n = -81 → -3n = -105 → n = 35.

Thus, -81 is the 35th term of the AP.

Is 0 a term?

Set 24 - 3n = 0 → -3n = -24 → n = 8.

Since 8 is a natural number, 0 is a term of the AP (the 8th term).

Q3. Exercise Set 8.2, Question 3 (Page 185): Find the n^th term of the AP: 11, 8, 5, 2, ... Write the recursive rule for this AP.

Answer:

The AP is 11, 8, 5, 2, ... with first term a = 11 and common difference d = 8 - 11 = -3.

The explicit formula for the n^th term is:

t_n = a + (n-1)d

= 11 + (n-1)(-3)

= 11 - 3n + 3

= 14 - 3n.

The recursive rule for the AP is:

t₁ = 11, t_n = t_n-1 - 3 for n ≥ 2.

Q4. Exercise Set 8.2, Question 4 (Page 185): 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'.)

Answer:

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

Given: t₃ = a + 2d = 12 ... (Equation 1)

The last term (50th term) is t₅₀ = a + 49d = 106 ... (Equation 2)

Subtracting Equation 1 from Equation 2:

(a + 49d) - (a + 2d)

= 106 - 12

 → 47d = 94

→ d = 2.

Substitute d = 2 in Equation 1: a + 2(2) = 12 → a + 4 = 12 → a = 8.

Now, the 29^th term is:

t₂₉ = a + 28d

= 8 + 28(2)

= 8 + 56 = 64.

Thus, the 29^th term is 64.

Q5. Exercise Set 8.2, Question 5 (Page 186): How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Answer:

Finding the 2-digit numbers divisible by 3:

The smallest 2-digit number divisible by 3 is 12. The largest is 99.

These numbers form an AP: 12, 15, 18, ..., 99 with first term a = 12 and common difference d = 3.

Let n be the number of terms.

 We have t_n = a + (n-1)d = 99.

12 + (n-1)3 = 99

→ (n-1)3 = 87

→ n-1 = 29

→ n = 30.

So, there are 30 two-digit numbers divisible by 3.

Finding the sum of these numbers:

The sum of an AP is given by S_n = n/2 × (first term + last term).

S = 30/2 × (12 + 99) = 15 × 111 = 1665.

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

Q6. Exercise Set 8.2, Question 6 (Page 186): 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?

Answer:

Let the number of years be n.

Harish's salary over the years forms an AP: ₹5,00,000; ₹5,20,000; ₹5,40,000; ...

Here, first term a = 5,00,000, common difference d = 20,000.

We need to find n such that the salary in the n^th year is ₹7,00,000.

t_n = a + (n-1)d = 7,00,000

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

(n-1)20,000 = 2,00,000 → n-1 = 10 → n = 11.

Thus, after 11 years, Harish's income will reach ₹7,00,000.

Q7. Exercise Set 8.2, Question 7 (Page 186): 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?

Answer:

The number of marbles in each row forms the sequence of natural numbers: 1, 2, 3, ..., up to 25 rows.

We need to find the sum of the first 25 natural numbers.

Using the formula S_n = n(n+1)/2:

S₂₅ = 25(26)/2 = 25 × 13 = 325.

Thus, the child uses 325 marbles in total.