GSEB Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1. Fill in the blanks in the following table. Given that \( a \) is the first term, \( d \) the common difference and \( a_n \) the nth term of the AP.

(i) Given: \( a = 7 \), \( d = 3 \), \( n = 8 \). We need to find \( a_n \).
The formula for the \( n \)th term of an AP is \( a_n = a + (n - 1)d \).
Substitute the known values into the formula:
\( a_n = 7 + (8 - 1) \times 3 \)
\( a_n = 7 + 7 \times 3 \)
\( a_n = 7 + 21 \)
\( a_n = 28 \)
So, the missing term is 28.

(ii) Given: \( a = -18 \), \( n = 10 \), \( a_n = 0 \). We need to find \( d \).
Use the formula \( a_n = a + (n - 1)d \).
\( 0 = -18 + (10 - 1)d \)
\( 0 = -18 + 9d \)
Move -18 to the other side:
\( 18 = 9d \)
Divide by 9 to get \( d \):
\( d = \frac{18}{9} \)
\( d = 2 \)
Thus, the common difference is 2.

(iii) Given: \( d = -3 \), \( n = 18 \), \( a_n = -5 \). We need to find \( a \).
Apply the formula \( a_n = a + (n - 1)d \).
\( -5 = a + (18 - 1)(-3) \)
\( -5 = a + (17)(-3) \)
\( -5 = a - 51 \)
Move -51 to the left side:
\( 51 - 5 = a \)
\( a = 46 \)
Hence, the first term is 46.

(iv) Given: \( a = -18.9 \), \( d = 2.5 \), \( a_n = 3.6 \). We need to find \( n \).
Use the formula \( a_n = a + (n - 1)d \).
\( 3.6 = -18.9 + (n - 1) \times 2.5 \)
Move -18.9 to the left side:
\( 3.6 + 18.9 = (n - 1) \times 2.5 \)
\( 22.5 = (n - 1) \times 2.5 \)
Divide both sides by 2.5:
\( \frac{22.5}{2.5} = n - 1 \)
\( 9 = n - 1 \)
Move -1 to the left side:
\( n = 9 + 1 \)
\( n = 10 \)
Therefore, the term number is 10.

(v) Given: \( a = 3.5 \), \( d = 0 \), \( n = 105 \). We need to find \( a_n \).
Apply the formula \( a_n = a + (n - 1)d \).
\( a_n = 3.5 + (105 - 1) \times 0 \)
Since any number multiplied by 0 is 0:
\( a_n = 3.5 + 0 \)
\( a_n = 3.5 \)
So, the missing term is 3.5.
In simple words: We used the main formula for arithmetic progression, \( a_n = a + (n - 1)d \), to find the unknown value in each part. We put in the numbers we knew and solved for the one we didn't.

Exam Tip: Remember the formula for the nth term of an AP: \( a_n = a + (n - 1)d \). Be careful with signs (positive/negative) and order of operations (multiplication before addition/subtraction).

Question 2. Choose the correct choice in the following and justify.
(i) 30th term of the AP: 10, 7, 4,..... is
(A) 97
(B) 77
(C) – 77
(D) – 87
Answer: (C) – 77
The given AP is 10, 7, 4, ...
Here, the first term \( a = 10 \).
The common difference \( d = 7 - 10 = -3 \).
We need to find the 30th term, so \( n = 30 \).
Using the formula for the nth term, \( a_n = a + (n - 1)d \):
\( a_{30} = 10 + (30 - 1)(-3) \)
\( a_{30} = 10 + (29)(-3) \)
\( a_{30} = 10 - 87 \)
\( a_{30} = -77 \)
Thus, the correct option is (C).
In simple words: For this list of numbers, the first number is 10 and each number goes down by 3. To find the 30th number, we start at 10 and subtract 3 for 29 times, which gives us -77.

Exam Tip: Always identify the first term \( (a) \), common difference \( (d) \), and the term number \( (n) \) before applying the formula \( a_n = a + (n - 1)d \). Pay close attention to negative signs when calculating the common difference.

(ii) 11th term of the AP: – 3, \( -\frac { 1 }{ 2 } \), 2, ..... is
(A) 28
(B) 22
(C) – 38
(D) 48
Answer: (B) 22
The given AP is \( -3, -\frac{1}{2}, 2, \ldots \)
Here, the first term \( a = -3 \).
The common difference \( d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = -\frac{1}{2} + \frac{6}{2} = \frac{5}{2} \).
We need to find the 11th term, so \( n = 11 \).
Using the formula for the nth term, \( a_n = a + (n - 1)d \):
\( a_{11} = -3 + (11 - 1) \times \frac{5}{2} \)
\( a_{11} = -3 + 10 \times \frac{5}{2} \)
\( a_{11} = -3 + 5 \times 5 \)
\( a_{11} = -3 + 25 \)
\( a_{11} = 22 \)
Hence, the correct option is (B).
In simple words: The first number in this series is -3, and each step adds \( \frac{5}{2} \) (which is 2.5). To get the 11th number, we start from -3 and add 2.5 ten times.

Exam Tip: When dealing with fractions in the common difference, ensure you perform the subtraction correctly by finding a common denominator. This helps prevent calculation errors.

Question 3. In the following APs, find the missing terms in the boxes.
(j) 2,      , 26
Answer:
(j) Let the common difference of the given AP be \( d \).
We have the first term \( a = 2 \).
The third term \( a_3 = 26 \).
Using the formula \( a_n = a + (n - 1)d \), for \( n=3 \):
\( a_3 = a + (3 - 1)d \)
\( 26 = 2 + 2d \)
Subtract 2 from both sides:
\( 26 - 2 = 2d \)
\( 24 = 2d \)
Divide by 2:
\( d = \frac{24}{2} \)
\( d = 12 \)
The second term \( a_2 \) is found using \( a_2 = a + d \).
\( a_2 = 2 + 12 \)
\( a_2 = 14 \)
So, the missing term in the box is 14.

(ii)      , 13,      , 3
Answer:
(ii) Let \( a \) be the first term and \( d \) be the common difference of the given AP.
We have the second term \( a_2 = 13 \)
\( a + d = 13 \) ...(1)
We also have the fourth term \( a_4 = 3 \)
\( a + 3d = 3 \) ...(2)
Subtract equation (1) from equation (2):
\( (a + 3d) - (a + d) = 3 - 13 \)
\( 2d = -10 \)
\( d = \frac{-10}{2} \)
\( d = -5 \)
Now, substitute the value of \( d \) into equation (1):
\( a + (-5) = 13 \)
\( a - 5 = 13 \)
\( a = 13 + 5 \)
\( a = 18 \)
So, the first missing term is 18.
Now find the third term \( a_3 \):
\( a_3 = a + 2d \)
\( a_3 = 18 + 2(-5) \)
\( a_3 = 18 - 10 \)
\( a_3 = 8 \)
Hence, the missing terms are 18 and 8.
In simple words: We used the given terms to create two equations with 'a' (first term) and 'd' (common difference). By solving these equations, we found 'a' and 'd', then calculated the missing terms.

(iii) 5,      ,      , \( 9\frac{1}{2} \)
Answer:
(iii) Let \( d \) be the common difference of the given AP.
The first term \( a = 5 \).
The fourth term \( a_4 = 9\frac{1}{2} \), which is \( \frac{19}{2} \).
Using the formula \( a_n = a + (n - 1)d \), for \( n=4 \):
\( a_4 = a + (4 - 1)d \)
\( \frac{19}{2} = 5 + 3d \)
Subtract 5 from both sides:
\( \frac{19}{2} - 5 = 3d \)
\( \frac{19}{2} - \frac{10}{2} = 3d \)
\( \frac{9}{2} = 3d \)
Divide by 3:
\( d = \frac{9}{2 \times 3} \)
\( d = \frac{9}{6} \)
\( d = \frac{3}{2} \)
Now, find the second term \( a_2 \):
\( a_2 = a + d = 5 + \frac{3}{2} \)
\( a_2 = \frac{10}{2} + \frac{3}{2} = \frac{13}{2} = 6\frac{1}{2} \)
Next, find the third term \( a_3 \):
\( a_3 = a + 2d = 5 + 2 \times \frac{3}{2} \)
\( a_3 = 5 + 3 \)
\( a_3 = 8 \)
Hence, the missing terms are \( 6\frac{1}{2} \) and 8.
In simple words: With the first term and the fourth term given, we calculated the common difference 'd'. Then, we used 'a' and 'd' to fill in the second and third missing numbers in the sequence.

(iv) – 4,      ,      ,      ,      , 6
Answer:
(iv) Let \( d \) be the common difference.
The first term \( a = -4 \).
The sixth term \( a_6 = 6 \).
Using the formula \( a_n = a + (n - 1)d \), for \( n=6 \):
\( a_6 = a + (6 - 1)d \)
\( 6 = -4 + 5d \)
Add 4 to both sides:
\( 6 + 4 = 5d \)
\( 10 = 5d \)
Divide by 5:
\( d = \frac{10}{5} \)
\( d = 2 \)
Now, find the missing terms:
Second term \( a_2 = a + d = -4 + 2 = -2 \)
Third term \( a_3 = a + 2d = -4 + 2(2) = -4 + 4 = 0 \)
Fourth term \( a_4 = a + 3d = -4 + 3(2) = -4 + 6 = 2 \)
Fifth term \( a_5 = a + 4d = -4 + 4(2) = -4 + 8 = 4 \)
Hence, the missing terms are -2, 0, 2, and 4.
In simple words: With the first and sixth terms given, we figured out that 'd' (the common step between numbers) is 2. Then, we just added 2 repeatedly to find all the numbers in between.

(v)      , 38,      ,      ,      , -22
Answer:
(v) Let \( a \) be the first term and \( d \) be the common difference.
We have the second term \( a_2 = 38 \)
\( a + d = 38 \) ...(1)
We also have the sixth term \( a_6 = -22 \)
\( a + 5d = -22 \) ...(2)
Subtract equation (1) from equation (2):
\( (a + 5d) - (a + d) = -22 - 38 \)
\( 4d = -60 \)
\( d = \frac{-60}{4} \)
\( d = -15 \)
Now, substitute the value of \( d \) into equation (1):
\( a + (-15) = 38 \)
\( a - 15 = 38 \)
\( a = 38 + 15 \)
\( a = 53 \)
So, the first missing term is 53.
Now find the third term \( a_3 \):
\( a_3 = a + 2d = 53 + 2(-15) \)
\( a_3 = 53 - 30 \)
\( a_3 = 23 \)
Next, find the fourth term \( a_4 \):
\( a_4 = a + 3d = 53 + 3(-15) \)
\( a_4 = 53 - 45 \)
\( a_4 = 8 \)
Finally, find the fifth term \( a_5 \):
\( a_5 = a + 4d = 53 + 4(-15) \)
\( a_5 = 53 - 60 \)
\( a_5 = -7 \)
Hence, the missing terms are 53, 23, 8, and -7.
In simple words: We used the second and sixth terms to create a system of equations, which helped us find the first term 'a' and the common difference 'd'. After getting 'a' and 'd', we could easily fill in all the blank spots in the sequence.

Exam Tip: For problems involving multiple missing terms, setting up simultaneous equations using the formula \( a_n = a + (n - 1)d \) is often the most effective method. Carefully solve for \( a \) and \( d \), then calculate each required term.

Question 4. Which term of the AP 3, 8, 13, 18, ....... is 78 ?
Answer: The given AP is 3, 8, 13, 18, ...
The first term is \( a = 3 \).
The common difference is \( d = 8 - 3 = 5 \).
We want to find which term is 78, so let \( a_n = 78 \).
Using the formula for the nth term, \( a_n = a + (n - 1)d \):
\( 78 = 3 + (n - 1) \times 5 \)
Subtract 3 from both sides:
\( 78 - 3 = (n - 1) \times 5 \)
\( 75 = (n - 1) \times 5 \)
Divide both sides by 5:
\( \frac{75}{5} = n - 1 \)
\( 15 = n - 1 \)
Add 1 to both sides:
\( n = 15 + 1 \)
\( n = 16 \)
Hence, the 16th term of the given AP is 78.
In simple words: We found that the sequence starts at 3 and increases by 5 each time. To reach 78, we worked backward using the formula, discovering that it is the 16th number in the sequence.

Exam Tip: When asked "Which term is...", you are looking for \( n \). Set up the equation \( a_n = a + (n - 1)d \) and solve for \( n \). Ensure your calculations for \( d \) and solving the linear equation are correct.

Question 5. Find the number of terms in each of the following APs.
(i) 7, 13, 19, ..., 205
Answer:
(i) The given AP is 7, 13, 19, ..., 205.
The first term is \( a = 7 \).
The common difference is \( d = 13 - 7 = 6 \).
The last term (nth term) is \( a_n = 205 \).
We need to find the number of terms, \( n \).
Using the formula \( a_n = a + (n - 1)d \):
\( 205 = 7 + (n - 1) \times 6 \)
Subtract 7 from both sides:
\( 205 - 7 = (n - 1) \times 6 \)
\( 198 = (n - 1) \times 6 \)
Divide both sides by 6:
\( \frac{198}{6} = n - 1 \)
\( 33 = n - 1 \)
Add 1 to both sides:
\( n = 33 + 1 \)
\( n = 34 \)
So, there are 34 terms in this arithmetic progression.
In simple words: This list of numbers starts at 7, ends at 205, and each step adds 6. We used a formula to count how many steps (terms) it takes to get from 7 to 205.

(ii) 18, \( 15\frac{1}{2} \), 13,...,- 47
Answer:
(ii) The given AP is 18, \( 15\frac{1}{2} \), 13, ..., -47.
The first term is \( a = 18 \).
The common difference is \( d = 15\frac{1}{2} - 18 = \frac{31}{2} - \frac{36}{2} = -\frac{5}{2} \).
The last term (nth term) is \( a_n = -47 \).
We need to find the number of terms, \( n \).
Using the formula \( a_n = a + (n - 1)d \):
\( -47 = 18 + (n - 1) \times \left(-\frac{5}{2}\right) \)
Subtract 18 from both sides:
\( -47 - 18 = (n - 1) \times \left(-\frac{5}{2}\right) \)
\( -65 = (n - 1) \times \left(-\frac{5}{2}\right) \)
Multiply both sides by \( -\frac{2}{5} \):
\( -65 \times \left(-\frac{2}{5}\right) = n - 1 \)
\( \frac{130}{5} = n - 1 \)
\( 26 = n - 1 \)
Add 1 to both sides:
\( n = 26 + 1 \)
\( n = 27 \)
Thus, there are 27 terms in this arithmetic progression.
In simple words: This number sequence starts at 18 and decreases by 2.5 each time, ending at -47. We calculated how many numbers are in this decreasing sequence.

Exam Tip: When finding the number of terms \( n \), ensure \( n \) is always a positive integer. If you get a fractional or negative value for \( n \), recheck your calculations. Be careful with calculations involving fractions and negative numbers for the common difference.

Question 6. Check whether – 150 is a term of the AP: 11, 8, 5, 2, ..........
Answer: The given list of numbers is 11, 8, 5, 2, ...
The first term is \( a = 11 \).
The common difference is \( d = 8 - 11 = -3 \).
Let's assume -150 is the \( n \)th term of this AP, so \( a_n = -150 \).
Using the formula \( a_n = a + (n - 1)d \):
\( -150 = 11 + (n - 1)(-3) \)
Subtract 11 from both sides:
\( -150 - 11 = (n - 1)(-3) \)
\( -161 = -3(n - 1) \)
Divide both sides by -3:
\( \frac{-161}{-3} = n - 1 \)
\( \frac{161}{3} = n - 1 \)
Add 1 to both sides:
\( n = \frac{161}{3} + 1 \)
\( n = \frac{161}{3} + \frac{3}{3} \)
\( n = \frac{164}{3} \)
Since \( n \) must be a positive integer (whole number) for a term in an AP, and \( \frac{164}{3} \) is not a whole number, -150 is not a term of the given AP.
In simple words: We checked if -150 could be a number in this sequence. Since we found that it would have to be term number \( \frac{164}{3} \), and term numbers must be whole numbers, -150 is not part of this list.

Exam Tip: For a number to be a term in an AP, the calculated value of \( n \) must always be a positive integer. If \( n \) is fractional or negative, the number is not part of the sequence.

Question 7. Find the 31st term of AP whose 11th term is 38 and 16th term is 73.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
We are given that the 11th term \( a_{11} = 38 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{11} = a + (11 - 1)d \)
\( a + 10d = 38 \) ...(1)
We are also given that the 16th term \( a_{16} = 73 \).
\( a_{16} = a + (16 - 1)d \)
\( a + 15d = 73 \) ...(2)
Now, subtract equation (1) from equation (2):
\( (a + 15d) - (a + 10d) = 73 - 38 \)
\( 5d = 35 \)
Divide by 5:
\( d = \frac{35}{5} \)
\( d = 7 \)
Substitute the value of \( d = 7 \) into equation (1):
\( a + 10(7) = 38 \)
\( a + 70 = 38 \)
Subtract 70 from both sides:
\( a = 38 - 70 \)
\( a = -32 \)
Now that we have \( a = -32 \) and \( d = 7 \), we can find the 31st term, \( a_{31} \).
\( a_{31} = a + (31 - 1)d \)
\( a_{31} = -32 + (30)(7) \)
\( a_{31} = -32 + 210 \)
\( a_{31} = 178 \)
Therefore, the 31st term of the AP is 178.
In simple words: We used the information about the 11th and 16th terms to find the starting number 'a' and how much the sequence increases by 'd'. Once we knew 'a' and 'd', we could easily figure out the 31st number in the sequence.

Exam Tip: When given two terms of an AP, form two linear equations in \( a \) and \( d \). Solve these simultaneous equations to find \( a \) and \( d \), then use these values to find any other required term.

Question 8. An AP consists of 50 terms of which third term is 12 and last term is 106. Find the 29th terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
The AP has a total of 50 terms, so \( n=50 \).
We are given that the third term \( a_3 = 12 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_3 = a + (3 - 1)d \)
\( a + 2d = 12 \) ...(1)
We are also given that the last term is 106. Since there are 50 terms, the last term is \( a_{50} \).
\( a_{50} = 106 \)
\( a_{50} = a + (50 - 1)d \)
\( a + 49d = 106 \) ...(2)
Now, subtract equation (1) from equation (2):
\( (a + 49d) - (a + 2d) = 106 - 12 \)
\( 47d = 94 \)
Divide by 47:
\( d = \frac{94}{47} \)
\( d = 2 \)
Substitute the value of \( d = 2 \) into equation (1):
\( a + 2(2) = 12 \)
\( a + 4 = 12 \)
Subtract 4 from both sides:
\( a = 12 - 4 \)
\( a = 8 \)
Now that we have \( a = 8 \) and \( d = 2 \), we can find the 29th term, \( a_{29} \).
\( a_{29} = a + (29 - 1)d \)
\( a_{29} = 8 + (28)(2) \)
\( a_{29} = 8 + 56 \)
\( a_{29} = 64 \)
Therefore, the 29th term is 64.
In simple words: Given the third term and the last (50th) term, we first found the starting number 'a' and the common step 'd'. Then, we used these to calculate the 29th number in the sequence.

Exam Tip: Remember that the "last term" is equivalent to \( a_n \) where \( n \) is the total number of terms. Constructing and solving simultaneous equations for \( a \) and \( d \) is crucial for these types of problems.

Question 9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
We are given that the 3rd term \( a_3 = 4 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_3 = a + (3 - 1)d \)
\( a + 2d = 4 \) ...(1)
We are also given that the 9th term \( a_9 = -8 \).
\( a_9 = a + (9 - 1)d \)
\( a + 8d = -8 \) ...(2)
Now, subtract equation (1) from equation (2):
\( (a + 8d) - (a + 2d) = -8 - 4 \)
\( 6d = -12 \)
Divide by 6:
\( d = \frac{-12}{6} \)
\( d = -2 \)
Substitute the value of \( d = -2 \) into equation (1):
\( a + 2(-2) = 4 \)
\( a - 4 = 4 \)
Add 4 to both sides:
\( a = 4 + 4 \)
\( a = 8 \)
Now we need to find which term is zero. Let \( a_n = 0 \).
Using the formula \( a_n = a + (n - 1)d \):
\( 0 = 8 + (n - 1)(-2) \)
Subtract 8 from both sides:
\( -8 = (n - 1)(-2) \)
Divide both sides by -2:
\( \frac{-8}{-2} = n - 1 \)
\( 4 = n - 1 \)
Add 1 to both sides:
\( n = 4 + 1 \)
\( n = 5 \)
Hence, the 5th term of the AP is zero.
In simple words: We used the given 3rd and 9th terms to figure out the starting number 'a' and the decrease 'd' in the sequence. Then, we used these to find which term in the sequence would be zero.

Exam Tip: This problem involves a two-step process: first, find \( a \) and \( d \) using the given terms, and then use these values to find the term number \( n \) for a specific value \( a_n \). Always check that \( n \) is a positive integer.

Question 10. The 17th term of an AP exceeds 10th term by 7. Find the common difference.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
The 17th term is \( a_{17} \). Using the formula \( a_n = a + (n - 1)d \):
\( a_{17} = a + (17 - 1)d \)
\( a_{17} = a + 16d \)
The 10th term is \( a_{10} \).
\( a_{10} = a + (10 - 1)d \)
\( a_{10} = a + 9d \)
According to the problem, the 17th term exceeds the 10th term by 7.
This means: \( a_{17} = a_{10} + 7 \)
Substitute the expressions for \( a_{17} \) and \( a_{10} \):
\( a + 16d = (a + 9d) + 7 \)
Subtract \( a \) from both sides:
\( 16d = 9d + 7 \)
Subtract \( 9d \) from both sides:
\( 16d - 9d = 7 \)
\( 7d = 7 \)
Divide by 7:
\( d = \frac{7}{7} \)
\( d = 1 \)
Thus, the common difference is 1.
In simple words: We know the 17th number in the sequence is 7 more than the 10th number. By setting up an equation with the first term 'a' and common difference 'd', we could find that the 'd' (the step size) is 1.

Exam Tip: When given relationships between terms (e.g., "exceeds by", "is less than"), always translate these into algebraic equations using the \( a_n = a + (n - 1)d \) formula. This allows you to solve for the unknown variables like \( d \) or \( a \).

Question 11. Which term of the AP: 3, 15, 27, 39, ..... will be 132 more than its 84th term?
Answer: The given AP is 3, 15, 27, 39, ...
The first term is \( a = 3 \).
The common difference is \( d = 15 - 3 = 12 \).
First, let's find the 84th term, \( a_{84} \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{84} = 3 + (84 - 1) \times 12 \)
\( a_{84} = 3 + (83) \times 12 \)
\( a_{84} = 3 + 996 \)
\( a_{84} = 999 \)
Now, we need to find which term \( (a_n) \) will be 132 more than its 84th term.
So, \( a_n = a_{84} + 132 \)
\( a_n = 999 + 132 \)
\( a_n = 1131 \)
Now we need to find the value of \( n \) for which \( a_n = 1131 \).
Using the formula \( a_n = a + (n - 1)d \):
\( 1131 = 3 + (n - 1) \times 12 \)
Subtract 3 from both sides:
\( 1131 - 3 = (n - 1) \times 12 \)
\( 1128 = (n - 1) \times 12 \)
Divide both sides by 12:
\( \frac{1128}{12} = n - 1 \)
\( 94 = n - 1 \)
Add 1 to both sides:
\( n = 94 + 1 \)
\( n = 95 \)
Hence, the 95th term of the AP will be 132 more than its 84th term.
In simple words: We first found the 84th number in the sequence. Then, we added 132 to it to get the target number. Finally, we used the AP formula to determine that this target number is the 95th term in the sequence.

Exam Tip: Break down complex problems into smaller steps. First, calculate the specified term (e.g., 84th term). Second, find the target value based on the given condition (132 more). Third, use the target value as \( a_n \) to solve for \( n \).

Question 12. Two APs have the same common difference. The difference between their 100th term is 100. What is the difference between 1000th terms?
Answer: Let the first AP have its first term as \( a_1 \) and the second AP have its first term as \( a_2 \).
Both APs have the same common difference, let's call it \( d \).
For the first AP, the 100th term is \( a_{1,100} = a_1 + (100 - 1)d = a_1 + 99d \).
For the second AP, the 100th term is \( a_{2,100} = a_2 + (100 - 1)d = a_2 + 99d \).
We are given that the difference between their 100th terms is 100.
So, \( a_{1,100} - a_{2,100} = 100 \)
\( (a_1 + 99d) - (a_2 + 99d) = 100 \)
\( a_1 + 99d - a_2 - 99d = 100 \)
\( a_1 - a_2 = 100 \)
Now, we need to find the difference between their 1000th terms.
For the first AP, the 1000th term is \( a_{1,1000} = a_1 + (1000 - 1)d = a_1 + 999d \).
For the second AP, the 1000th term is \( a_{2,1000} = a_2 + (1000 - 1)d = a_2 + 999d \).
The difference between their 1000th terms is:
\( a_{1,1000} - a_{2,1000} = (a_1 + 999d) - (a_2 + 999d) \)
\( = a_1 + 999d - a_2 - 999d \)
\( = a_1 - a_2 \)
Since we already found that \( a_1 - a_2 = 100 \),
The difference between their 1000th terms is also 100.
In simple words: Because both sequences increase by the same amount 'd' at each step, the difference between any two corresponding terms (like the 100th terms or 1000th terms) will always be the same as the difference between their first terms. Since the 100th terms differ by 100, the 1000th terms will also differ by 100.

Exam Tip: A key property of APs with the same common difference is that the difference between any two corresponding terms is constant and equal to the difference between their first terms. This understanding can simplify such problems significantly.

Question 13. How many three-digit numbers are divisible by 7 ?
Answer: We need to find the three-digit numbers that are divisible by 7.
The smallest three-digit number is 100. Let's find the first three-digit number divisible by 7.
\( 100 \div 7 \approx 14.28 \). The next multiple of 7 after \( 14 \times 7 = 98 \) is \( 15 \times 7 = 105 \).
So, the first three-digit number divisible by 7 is \( a = 105 \).
The largest three-digit number is 999. Let's find the last three-digit number divisible by 7.
\( 999 \div 7 \approx 142.71 \). So, \( 142 \times 7 = 994 \).
The last three-digit number divisible by 7 is \( a_n = 994 \).
These numbers form an AP with first term \( a = 105 \) and common difference \( d = 7 \).
We need to find the number of terms, \( n \).
Using the formula \( a_n = a + (n - 1)d \):
\( 994 = 105 + (n - 1) \times 7 \)
Subtract 105 from both sides:
\( 994 - 105 = (n - 1) \times 7 \)
\( 889 = (n - 1) \times 7 \)
Divide both sides by 7:
\( \frac{889}{7} = n - 1 \)
\( 127 = n - 1 \)
Add 1 to both sides:
\( n = 127 + 1 \)
\( n = 128 \)
Hence, there are 128 three-digit numbers that are divisible by 7.
In simple words: We found the first three-digit number that 7 divides evenly into (105) and the last one (994). Then, we used the formula for arithmetic progression to count how many such numbers exist between them.

Exam Tip: To find the first term \( a \), divide the smallest number in the range by the divisor and round up to the next integer, then multiply by the divisor. To find the last term \( a_n \), divide the largest number in the range by the divisor and round down to the nearest integer, then multiply by the divisor.

Question 14. How many multiples of 4 lie between 10 and 250?
Answer: We need to find the multiples of 4 that are between 10 and 250.
The numbers "between 10 and 250" mean we consider numbers strictly greater than 10 and strictly less than 250.
The first multiple of 4 greater than 10 is 12. (Since \( 10 \div 4 = 2.5 \), the next multiple is \( 3 \times 4 = 12 \)).
So, the first term of our AP is \( a = 12 \).
The last multiple of 4 less than 250 is 248. (Since \( 250 \div 4 = 62.5 \), the largest multiple is \( 62 \times 4 = 248 \)).
So, the last term of our AP is \( a_n = 248 \).
The common difference is \( d = 4 \) (since we are looking for multiples of 4).
We need to find the number of terms, \( n \).
Using the formula \( a_n = a + (n - 1)d \):
\( 248 = 12 + (n - 1) \times 4 \)
Subtract 12 from both sides:
\( 248 - 12 = (n - 1) \times 4 \)
\( 236 = (n - 1) \times 4 \)
Divide both sides by 4:
\( \frac{236}{4} = n - 1 \)
\( 59 = n - 1 \)
Add 1 to both sides:
\( n = 59 + 1 \)
\( n = 60 \)
Hence, there are 60 multiples of 4 between 10 and 250.
In simple words: We first identified the smallest multiple of 4 after 10 (which is 12) and the largest multiple of 4 before 250 (which is 248). Then, using the arithmetic progression formula, we counted how many such multiples are in that range.

Exam Tip: Pay close attention to keywords like "between" (exclusive) versus "from...to" (inclusive) when determining the first and last terms of the sequence. For "between X and Y", X and Y themselves are not included.

Question 18. The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three term of the AP.
Answer: Let 'a' be the first term and 'd' be the common difference of the Arithmetic Progression (AP).
The 4th term is \( a_4 = a + 3d \).
The 8th term is \( a_8 = a + 7d \).
According to the problem, the sum of the 4th and 8th terms is 24:
\( a_4 + a_8 = 24 \)
\( (a + 3d) + (a + 7d) = 24 \)
\( 2a + 10d = 24 \)
Dividing by 2, we get:
\( a + 5d = 12 \) ......(1)
The 6th term is \( a_6 = a + 5d \).
The 10th term is \( a_{10} = a + 9d \).
According to the problem, the sum of the 6th and 10th terms is 44:
\( a_6 + a_{10} = 44 \)
\( (a + 5d) + (a + 9d) = 44 \)
\( 2a + 14d = 44 \)
Dividing by 2, we get:
\( a + 7d = 22 \) ......(2)
Now, we solve equations (1) and (2) simultaneously. Subtract equation (1) from equation (2):
\( (a + 7d) - (a + 5d) = 22 - 12 \)
\( 2d = 10 \)
\( d = 5 \)
Substitute the value of \( d \) into equation (1):
\( a + 5(5) = 12 \)
\( a + 25 = 12 \)
\( a = 12 - 25 \)
\( a = -13 \)
The first three terms of the AP are \( a \), \( a + d \), and \( a + 2d \):
First term: \( a = -13 \)
Second term: \( a + d = -13 + 5 = -8 \)
Third term: \( a + 2d = -13 + 2(5) = -13 + 10 = -3 \)
So, the first three terms are -13, -8, and -3.
In simple words: We used the given sums to create two equations with 'a' (first term) and 'd' (common difference). Solving these equations gave us 'a' as -13 and 'd' as 5. Then, we calculated the first three terms of the AP: -13, -8, and -3.

Exam Tip: For problems involving sums of terms, always write down the general formula for the nth term \( a_n = a + (n-1)d \) and form simultaneous equations to find 'a' and 'd'.

Question 19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer: This situation forms an Arithmetic Progression (AP).
The initial salary (first term) is \( a = \text{Rs. } 5000 \).
The annual increment (common difference) is \( d = \text{Rs. } 200 \).
The target income (nth term) is \( a_n = \text{Rs. } 7000 \).
We use the formula for the nth term of an AP: \( a_n = a + (n - 1)d \).
Substitute the given values into the formula:
\( 7000 = 5000 + (n - 1)200 \)
Subtract 5000 from both sides:
\( 7000 - 5000 = (n - 1)200 \)
\( 2000 = (n - 1)200 \)
Divide both sides by 200:
\( \frac{2000}{200} = n - 1 \)
\( 10 = n - 1 \)
Add 1 to both sides:
\( n = 10 + 1 \)
\( n = 11 \)
So, in the 11th year, Subba Rao's salary will become Rs. 7000.
In simple words: We set up an AP where the starting salary is the first term, the increment is the common difference, and the target salary is the nth term. By using the AP formula, we found that it took 11 years to reach Rs. 7000.

Exam Tip: Remember to clearly identify the first term (a), common difference (d), and the nth term (\(a_n\)) from the problem statement before applying the formula \(a_n = a + (n-1)d\).

Question 20. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week her weekly savings becomes Rs. 20.75, find n.
Answer: This problem also involves an Arithmetic Progression (AP).
The initial saving in the first week (first term) is \( a = \text{Rs. } 5 \).
The weekly increase in savings (common difference) is \( d = \text{Rs. } 1.75 \).
The saving in the nth week (nth term) is \( a_n = \text{Rs. } 20.75 \).
We use the formula for the nth term of an AP: \( a_n = a + (n - 1)d \).
Substitute the known values into the formula:
\( 20.75 = 5 + (n - 1)1.75 \)
Subtract 5 from both sides:
\( 20.75 - 5 = (n - 1)1.75 \)
\( 15.75 = (n - 1)1.75 \)
Divide both sides by 1.75:
\( \frac{15.75}{1.75} = n - 1 \)
To simplify the division, multiply the numerator and denominator by 100:
\( \frac{1575}{175} = n - 1 \)
\( 9 = n - 1 \)
Add 1 to both sides:
\( n = 9 + 1 \)
\( n = 10 \)
Therefore, in the 10th week, Ramkali's weekly savings will become Rs. 20.75.
In simple words: Ramkali's savings follow a pattern where she adds the same amount each week. We used the starting amount, the amount she adds weekly, and her target saving to figure out that it took 10 weeks to reach that target.

Exam Tip: Be careful with decimal calculations. You can remove decimals by multiplying both numerator and denominator by a suitable power of 10 to make the division simpler.