Exercise 8.2 Practice
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Overview
Solve arithmetic progression term formulas, linear growth models, and divisible integer counts. Practice finding specific terms and identifying common differences in structured mathematical sequences.
Arithmetic Progressions (AP), Common Differences, Sums & Practical Word Problems
Q1: AP Specific Terms
Find the 10th and 26th terms of the AP: $3, 8, 13, 18, \dots$.
Identify first term $a$ and common difference $d$ of the AP:
• $a = 3$
• $d = 8 - 3 = 5$
• $a = 3$
• $d = 8 - 3 = 5$
The general term is $t_n = a + (n - 1)d$:
• **10th term ($n = 10$):**
$$t_{10} = 3 + (10 - 1)5 = 3 + 9(5) = 3 + 45 = \mathbf{48}$$
• **26th term ($n = 26$):**
$$t_{26} = 3 + (26 - 1)5 = 3 + 25(5) = 3 + 125 = \mathbf{128}$$
• **10th term ($n = 10$):**
$$t_{10} = 3 + (10 - 1)5 = 3 + 9(5) = 3 + 45 = \mathbf{48}$$
• **26th term ($n = 26$):**
$$t_{26} = 3 + (26 - 1)5 = 3 + 25(5) = 3 + 125 = \mathbf{128}$$
10th term = 48, 26th term = 128
Q2: Position Check & Zero Analysis
Which term of the AP: $21, 18, 15, \dots$ is $-81$? Also, is $0$ a term of this AP? Give reasons for your answer.
Identify first term $a = 21$ and common difference $d = 18 - 21 = -3$.
Check for $-81$:
Set $t_n = a + (n - 1)d = -81$:
$$21 + (n - 1)(-3) = -81$$
$$(n - 1)(-3) = -81 - 21 = -102$$
$$n - 1 = \frac{-102}{-3} = 34 \implies n = 35$$
Since $35$ is a natural number, $-81$ is the 35th term of the AP.
Set $t_n = a + (n - 1)d = -81$:
$$21 + (n - 1)(-3) = -81$$
$$(n - 1)(-3) = -81 - 21 = -102$$
$$n - 1 = \frac{-102}{-3} = 34 \implies n = 35$$
Since $35$ is a natural number, $-81$ is the 35th term of the AP.
Check for $0$:
Set $t_n = a + (n - 1)d = 0$:
$$21 + (n - 1)(-3) = 0$$
$$(n - 1)(-3) = -21$$
$$n - 1 = \frac{-21}{-3} = 7 \implies n = 8$$
Since $8$ is a natural number, $0$ is the 8th term of the AP.
Set $t_n = a + (n - 1)d = 0$:
$$21 + (n - 1)(-3) = 0$$
$$(n - 1)(-3) = -21$$
$$n - 1 = \frac{-21}{-3} = 7 \implies n = 8$$
Since $8$ is a natural number, $0$ is the 8th term of the AP.
$-81$ is the 35th term; $0$ is the 8th term.
Q3: General Term & Recursive Rule
Find the $n$-th term of the AP: $11, 8, 5, 2, \dots$. Write the recursive rule for this AP.
Identify $a = 11$ and $d = 8 - 11 = -3$.
Explicit $n$-th Term Formula:
$$t_n = a + (n - 1)d = 11 + (n - 1)(-3) = 11 - 3n + 3 = \mathbf{14 - 3n}$$
$$t_n = a + (n - 1)d = 11 + (n - 1)(-3) = 11 - 3n + 3 = \mathbf{14 - 3n}$$
Recursive Rule:
$$t_1 = 11, \quad \mathbf{t_n = t_{n-1} - 3 \text{ for } n \ge 2}$$
$$t_1 = 11, \quad \mathbf{t_n = t_{n-1} - 3 \text{ for } n \ge 2}$$
Explicit: $t_n = 14 - 3n$; Recursive: $t_1 = 11, t_n = t_{n-1} - 3$
Q4: AP Equation Solver
An AP consists of $50\text{ terms}$ in which the 3rd term is $12$ and the last term is $106$. Find the 29th term.
We are given $t_3 = 12$ and last term $t_{50} = 106$.
We can represent these terms algebraically:
1. $a + 2d = 12$
2. $a + 49d = 106$
We can represent these terms algebraically:
1. $a + 2d = 12$
2. $a + 49d = 106$
Subtract Equation 1 from Equation 2 to eliminate $a$:
$$(a + 49d) - (a + 2d) = 106 - 12$$
$$47d = 94 \implies d = 2$$
$$(a + 49d) - (a + 2d) = 106 - 12$$
$$47d = 94 \implies d = 2$$
Substitute $d = 2$ back into Equation 1 to find $a$:
$$a + 2(2) = 12 \implies a + 4 = 12 \implies a = 8$$
$$a + 2(2) = 12 \implies a + 4 = 12 \implies a = 8$$
Now find the 29th term ($t_{29}$):
$$t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}$$
$$t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}$$
29th term = 64
Q5: Numbers Divisible by 3 & Sum
How many 2-digit numbers are divisible by $3$? What is the sum of all these 2-digit numbers?
Step 1: Identify the AP of 2-digit numbers divisible by 3:
The first 2-digit number divisible by 3 is $12$. The last is $99$.
Sequence: $12, 15, 18, \dots, 99$ (AP with $a = 12$, $d = 3$, $t_n = 99$).
The first 2-digit number divisible by 3 is $12$. The last is $99$.
Sequence: $12, 15, 18, \dots, 99$ (AP with $a = 12$, $d = 3$, $t_n = 99$).
Step 2: Find total number of terms ($n$):
$$t_n = a + (n - 1)d \implies 99 = 12 + (n - 1)3$$
$$87 = (n - 1)3 \implies n - 1 = 29 \implies n = \mathbf{30}$$
$$t_n = a + (n - 1)d \implies 99 = 12 + (n - 1)3$$
$$87 = (n - 1)3 \implies n - 1 = 29 \implies n = \mathbf{30}$$
Step 3: Calculate the sum ($S_{30}$):
$$S_n = \frac{n}{2}(a + t_n)$$
$$S_{30} = \frac{30}{2}(12 + 99) = 15(111) = \mathbf{1665}$$
$$S_n = \frac{n}{2}(a + t_n)$$
$$S_{30} = \frac{30}{2}(12 + 99) = 15(111) = \mathbf{1665}$$
There are 30 such numbers; Sum = 1665
Q6: Salary Increment AP Word Problem
Harish started work at an annual salary of $\text{₹}5,00,000$ and received an increment of $\text{₹}20,000$ each year. After how many years did his income reach $\text{₹}7,00,000$?
Let Harish's salary be represented by an AP:
• First term (initial salary) $a = 5,00,000$
• Common difference (annual increment) $d = 20,000$
• Target term $t_n = 7,00,000$
• First term (initial salary) $a = 5,00,000$
• Common difference (annual increment) $d = 20,000$
• Target term $t_n = 7,00,000$
Set up the $n$-th term equation:
$$t_n = a + (n - 1)d$$
$$7,00,000 = 5,00,000 + (n - 1)20,000$$
$$2,00,000 = (n - 1)20,000$$
$$n - 1 = \frac{2,00,000}{20,000} = 10 \implies n = 11$$
$$t_n = a + (n - 1)d$$
$$7,00,000 = 5,00,000 + (n - 1)20,000$$
$$2,00,000 = (n - 1)20,000$$
$$n - 1 = \frac{2,00,000}{20,000} = 10 \implies n = 11$$
The salary reaches ₹7,00,000 in the 11th year. The number of increments received is $n - 1 = 10$, which means it took **10 years** of service to reach this income.
Income reached ₹7,00,000 after 10 years (in the 11th year)
Q7: Marble Arrangement Natural Number Sum
A child arranges marbles in rows so that the first row has $1\text{ marble}$, the second has $2\text{ marbles}$, the third has $3$, and so on up to $25\text{ rows}$. How many marbles does the child use in all?
This represents the sum of the first 25 natural numbers:
$$\text{Total Marbles} = 1 + 2 + 3 + \dots + 25$$
$$\text{Total Marbles} = 1 + 2 + 3 + \dots + 25$$
Using the sum formula $S_n = \frac{n(n + 1)}{2}$ for $n = 25$:
$$S_{25} = \frac{25 \times 26}{2} = 25 \times 13 = \mathbf{325}$$
$$S_{25} = \frac{25 \times 26}{2} = 25 \times 13 = \mathbf{325}$$
The child uses 325 marbles in total