Welcome back to our math blog! Today, we are going to solve problems from Exercise Set 8.2.
If you have been following along in class, you know that an AP is just a sequence of numbers where the jump (or difference) between every consecutive term is exactly the same.
Q1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Solution
Identify the components of the given AP
First, let’s look at the sequence given to us: 3, 8, 13, 18, \dots.
First term, a = 3
Common d = 8 - 3 = 5
The formula for the n^{th} term (t_n) is:
t_n = a + (n - 1)d
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To find the 10th term, we simply plug n = 10 into our formula, along with our values for a and d.
a = 3
d = 5
n = 10
t_n = a + (n - 1)d
t_{10} = 3 + (10 - 1) \times 5
t_{10} = 3 + (9) \times 5
t_{10} = 3 + 45
t_{10} = 48
So, the 10th term of this AP is 48.
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Now, we repeat the exact same process to find the 26th term. This time, we substitute n = 26.
a = 3
d = 5
n = 26
t_n = a + (n - 1)d
t_{26} = 3 + (26 - 1) \times 5
t_{26} = 3 + (25) \times 5
t_{26} = 3 + 125
t_{26} = 128
So, the 26th term of this AP is 128.
Final Answer: \boxed {t_{10} = 48 \ \ t_{26} = 128}
Q2. Which term of the AP: 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution:
Identify the components of the given AP First, let’s look at the sequence given to us: 21, 18, 15, \dots.
First term, a = 21
Common difference, d = 18 - 21 = -3
The formula for the n^{th} term (t_n) is: t_n = a + (n - 1)d
To find which term is -81, we plug t_n = -81 into our formula, along with our values for a and d.
a = 21
d = -3
t_n = -81
t_n = a + (n - 1)d
-81 = 21 + (n - 1) \times (-3)
-81 - 21 = -3(n - 1)
-102 = -3(n - 1)
34 = n - 1
n = 35
So, the 35th term of this AP is -81.
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Now, we repeat the process to check if 0 is a term of this AP.
This time, we substitute t_n = 0.
a = 21
d = -3
t_n = 0
t_n = a + (n - 1)d
0 = 21 + (n - 1) \times (-3)
-21 = -3(n - 1)
7 = n - 1
n = 8
Since n is a positive integer, 0 is indeed a valid term in this sequence.
Final Answer: \boxed {\text{35th term is } -81, \text{ and } 0 \text{ is the 8th term}}
Q3. Find the n^{th} term of the AP: 11, 8, 5, 2… Write the recursive rule for this AP.
Solution:
Identify the components of the given AP First, let’s look at the sequence given to us: 11, 8, 5, 2, \dots.
First term, a = 11
Common difference, d = 8 - 11 = -3
The formula for the n^{th} term (t_n) is: t_n = a + (n - 1)d
To find the n^{th} term, we plug our values for a and d into the formula, leaving n as a variable.
a = 11
d = -3
t_n = a + (n - 1)d
t_n = 11 + (n - 1) \times (-3)
t_n = 11 - 3n + 3
t_n = 14 - 3n
So, the explicit rule for the n^{th} term is 14 - 3n.
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Now, we need to write the recursive rule for this AP.
A recursive formula gives the value of a given term using the values of previous terms.
t_1 = 11
Since the common difference is -3, each term is 3 less than the previous term.
t_n = t_{n-1} - 3
So, the recursive rule is t_1 = 11, t_n = t_{n-1} - 3 for n \ge 2.
Final Answer: \boxed {t_n = 14 - 3n \text{ and recursive rule: } t_1 = 11, t_n = t_{n-1} - 3}
Q4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Identify the components of the given AP Here, the sequence is not given directly, but we are given specific terms.
Total terms, n = 50
3rd term, t_3 = 12
Last term (50th term), t_{50} = 106
The formula for the n^{th} term (t_n) is: t_n = a + (n - 1)d
To find a and d, we create a pair of linear equations using the given terms.
For the 3rd term (n = 3):
t_3 = a + (3 -1)d
12 = a + 2d (Equation 1)
For the 50th term (n = 50):
t_{50} = a + (50 - 1)d
106 = a + 49d (Equation 2)
Now, we solve these two equations by subtracting Equation 1 from Equation 2.
(a + 49d) - (a + 2d) = 106 - 12
47d = 94
d = 2
Substitute d = 2 back into Equation 1 to find a.
12 = a + 2(2)
12 = a + 4
a = 8
So, the first term is 8 and the common difference is 2.
Now, we find the 29th term.
This time, we substitute n = 29.
a = 8
d = 2
n = 29
t_n = a + (n - 1)d
t_{29} = 8 + (29 - 1) \times 2
t_{29} = 8 + (28) \times 2
t_{29} = 8 + 56
t_{29} = 64
So, the 29th term of this AP is 64.
Final Answer: \boxed {t_{29} = 64}
Q5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution:
Identify the components of the given AP First,
let’s identify the sequence of 2-digit numbers divisible by 3.
The smallest is 12, and the largest is 99. Sequence: 12, 15, 18, \dots, 99
First term, a = 12
Common difference, d = 15 - 12 = 3
Last term, t_n = 99
The formula for the n^{th} term (t_n) is: t_n = a + (n - 1)d
To find how many numbers there are (n), we plug our values into the formula.
a = 12
d = 3
t_n = 99
t_n = a + (n - 1)d
99 = 12 + (n - 1) \times 3
99 - 12 = 3(n - 1)
87 = 3(n - 1)
29 = n - 1
n = 30
So, there are 30 two-digit numbers divisible by 3.
Now, we find the sum of these 30 numbers using the AP sum formula:
S_n = \frac{n}{2}(a + l), where l is the last term.
n = 30
a = 12
l = 99
S_{30} = \frac{30}{2}(12 + 99)
S_{30} = 15 \times 111
S_{30} = 1665
So, the sum of all these 2-digit numbers is 1665.
Final Answer: \boxed {n = 30, \text{ Sum} = 1665}
Q6. 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?
Solution:
Identify the components of the given AP First, let’s look at the sequence of Harish’s salary over the years.
Sequence: 500000, 520000, 540000, \dots, 700000
First term, a = 500000
Common difference, d = 20000
Target term, t_n = 700000
The formula for the n^{th} term (t_n) is: t_n = a + (n - 1)d
To find the year (n) his salary reaches ₹7,00,000, we plug our values into the formula.
a = 500000
d = 20000
t_n = 700000
t_n = a + (n - 1)d
700000 = 500000 + (n - 1) \times 20000
700000 - 500000 = 20000(n - 1)
200000 = 20000(n - 1)
10 = n - 1
n = 11
So, his income reached ₹7,00,000 in his 11th year of employment.
The question specifically asks “After how many years”. Since he reaches this salary in year 11, it means this happens after 10 full years of increments have been applied.
So, his income reaches ₹7,00,000 after 10 years.
Final Answer: \boxed {\text{After 10 years (In the 11th year)}}
Q7. 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?
Solution:
Identify the components of the given AP First, let’s look at the sequence representing the number of marbles in each row:
1, 2, 3, \dots, 25.
First term, a = 1
Common difference, d = 2 - 1 = 1
Total number of rows, n = 25
The formula for the sum of the first n natural numbers (S_n) is: S_n = \frac{n(n + 1)}{2}
To find the total number of marbles, we simply plug n = 25 into our formula.
n = 25
S_n = \frac{n(n + 1)}{2}
S_{25} = \frac{25(25 + 1)}{2}
S_{25} = \frac{25 \times 26}{2}
S_{25} = 25 \times 13
S_{25} = 325
So, the child uses 325 marbles in all.
Final Answer: \boxed {\text{Total marbles} = 325}