Exercise 8.2 Practice
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Overview
This page provides comprehensive Ch 8: Exploring Sequences and Progressions - Exercise Set 8.2 Practice. Practice solving Arithmetic Progression (AP) term formulas, sum calculations, linear growth real-life models, and divisible integer counts with step-by-step solutions.
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