CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 02

ARITHMETIC PROGRESSIONS

EASY SCORING IMPORTANT QUESTIONS

LEVEL- 1 (1 Marks)

Q1) For the A.P 3, 1, -1, -3...., Write the first term and common difference?
Answer: Here, first term \(a = 3\)
Common difference \(d = 1 - 3 = -2\)
In simple words: The first term is the starting number, and the common difference is found by subtracting any term from the one after it.

Exam Tip: Always subtract the first term from the second term, not the other way around, to get the correct sign for \(d\).

Q2) Find the 10th term of the A.P, whose first term is 2 and common difference is 5.
Answer: Here, first term \(a = 2\) and common difference \(d = 5\)
Formula: \(a_n = a + (n - 1) \times d\)
\(a_{10} = 2 + (10 - 1) \times 5\)
\(a_{10} = 2 + 9 \times 5\)
\(a_{10} = 2 + 45\)
\(a_{10} = 47\)
In simple words: Substitute \(n = 10\), \(a = 2\), and \(d = 5\) into the general term formula.

Exam Tip: Remember that the coefficient of \(d\) is always one less than the term number you want to find (\(n - 1\)).

Q3. Write the formula to find the sum of first \(n\) terms of an AP whose first term is \(a\) and common difference is \(d\).
Answer: \(S_n = \frac{n}{2} \{ 2a + (n - 1)d \}\)
In simple words: This is the standard formula used to find the sum of an arithmetic progression when the first term and common difference are known.

Exam Tip: Memorize this formula thoroughly as it is frequently applied in multi-mark questions.

Q4. Write the sum of first \(n\) terms of an AP whose first term is \(a\) and last term is \(l\).
Answer: \(S_n = \frac{n}{2} (a + l)\)
In simple words: Use this short formula when the last term of the AP is already given.

Exam Tip: This formula saves time when the last term \(l\) (or \(a_n\)) is provided in the problem statement.

LEVEL- 2 (2 Marks)

Q1. Find the 8th term of an A.P. 7, 10, 13, .....?
Answer: Here, first term \(a = 7\)
Common difference \(d = 10 - 7 = 3\)
Using the formula \(a_n = a + (n - 1)d\), for the 8th term:
\(a_8 = a + 7d\)
\(a_8 = 7 + 7 \times 3\)
\(a_8 = 7 + 21\)
\(a_8 = 28\)
In simple words: Find \(a\) and \(d\) from the given sequence, then substitute into the formula for \(a_8\).

Exam Tip: Writing \(a_8 = a + 7d\) directly skips one step and reduces calculation errors.

Q2. Find the sum of first 10 terms of the A.P 2, 5, 8, 11 .......?
Answer: Here, first term \(a = 2\)
Common difference \(d = 5 - 2 = 3\)
Number of terms \(n = 10\)
Using the sum formula \(S_n = \frac{n}{2} [ 2a + (n - 1)d ]\):
\(S_{10} = \frac{10}{2} [ 2 \times 2 + (10 - 1) \times 3 ]\)
\(S_{10} = 5 [ 4 + 9 \times 3 ]\)
\(S_{10} = 5 [ 4 + 27 ]\)
\(S_{10} = 5 \times 31\)
\(S_{10} = 155\)
In simple words: Identify \(a\), \(d\), and \(n\), plug them into the sum formula, and simplify carefully following order of operations.

Exam Tip: Always evaluate the inner bracket expression completely before multiplying by the outer factor.

Q3. For what values of \(n\) the \(n\)th term of AP is 63, 65, 67, .......... and 3, 10, 17, ...... are equal?
Answer: For AP 1: 63, 65, 67, ...
First term \(a = 63\), common difference \(d = 65 - 63 = 2\)
\(n\)th term of AP 1: \(a_n = a + (n - 1)d = 63 + (n - 1)2\)
For AP 2: 3, 10, 17, ...
First term \(A = 3\), common difference \(D = 10 - 3 = 7\)
\(n\)th term of AP 2: \(A_n = A + (n - 1)D = 3 + (n - 1)7\)
Equating the two \(n\)th terms: 
\(63 + (n - 1)2 = 3 + (n - 1)7\)
\(63 - 3 = (n - 1)7 - (n - 1)2\)
\(60 = (n - 1)(7 - 2)\)
\(60 = (n - 1)5\)
\(\frac{60}{5} = n - 1\)
\(12 = n - 1\)
\(n = 12 + 1 = 13\)
In simple words: Write expressions for the \(n\)th term of both progressions, set them equal to each other, and solve the linear equation for \(n\).

Exam Tip: Grouping terms containing \((n - 1)\) on one side avoids expanding polynomials and simplifies the algebra.

Q4. Find the 20th term from the last term of an AP 3, 8, 13, ...... 253.
Answer: Given AP: 3, 8, 13, ..., 253
Here, first term \(a = 3\), common difference \(d = 8 - 3 = 5\), and last term \(l = 253\)
To find terms from the end, reverse the AP where the new first term \(A = 253\) and the new common difference \(D = -5\)
Using the formula for the \(n\)th term: 
\(A_{20} = A + (20 - 1)D\)
\(A_{20} = 253 + 19 \times (-5)\)
\(A_{20} = 253 - 95\)
\(A_{20} = 158\)
In simple words: Reversing the AP turns the last term into the first term and changes the sign of the common difference.

Exam Tip: Remember to change the sign of \(d\) to its negative when measuring terms from the end of the sequence.

LEVEL- 3 (3 Marks)

Q1) Which term of the A.P : 21, 18, 15, ..... is -81?
Answer: Here, first term \(a = 21\)
Common difference \(d = 18 - 21 = -3\)
Let the \(n\)th term be \(-81\), so \(a_n = -81\)
\(a + (n - 1)d = -81\)
\(21 + (n - 1)(-3) = -81\)
\((n - 1)(-3) = -81 - 21\)
\((n - 1)(-3) = -102\)
\(n - 1 = \frac{-102}{-3}\)
\(n - 1 = 34\)
\(n = 34 + 1 = 35\)
Hence, the 35th term is -81.
In simple words: Equate the \(n\)th term formula to -81 and solve for \(n\) to find the position number.

Exam Tip: Ensure proper handling of negative signs during division to prevent sign errors.

Q2. Determine the A.P whose 3rd term is 5 and 7th term is 9?
Answer: Let the first term of the A.P be \(a\) and the common difference be \(d\).
Given, 3rd term \(a_3 = 5\)
\(a + (3 - 1)d = 5\)
\(a + 2d = 5 \text{ } ........ (1)\)
Also, 7th term \(a_7 = 9\)
\(a + (7 - 1)d = 9\)
\(a + 6d = 9 \text{ } ........ (2)\)
Subtracting equation (1) from equation (2):
\((a + 6d) - (a + 2d) = 9 - 5\)
\(4d = 4\)
\(d = 1\)
Putting \(d = 1\) in equation (1):
\(a + 2(1) = 5\)
\(a + 2 = 5\)
\(a = 5 - 2 = 3\)
Required A.P is \(a, a + d, a + 2d, a + 3d, ...\)
\(3, (3 + 1), (3 + 2 \times 1), (3 + 3 \times 1), ...\)
\(3, 4, 5, 6, ...\)
In simple words: Set up a system of two linear equations in terms of \(a\) and \(d\), solve for both variables, and build the sequence.

Exam Tip: Elimination by subtraction is the fastest way to solve simultaneous linear equations involving \(a\).

Q3. Find the common difference of an A.P. whose first term is \(\frac{1}{2}\) and the 8th term is \(\frac{17}{6}\). Also write its 4th term.
Answer: Here, first term \(a = \frac{1}{2}\)
Eighth term \(a_8 = \frac{17}{6}\)
\(a + 7d = \frac{17}{6}\)
Substitute \(a = \frac{1}{2}\):
\(\frac{1}{2} + 7d = \frac{17}{6}\)
\(7d = \frac{17}{6} - \frac{1}{2}\)
\(7d = \frac{17 - 3}{6} = \frac{14}{6} = \frac{7}{3}\)
\(d = \frac{7}{3 \times 7} = \frac{1}{3}\)
Now, find the 4th term \(a_4\):
\(a_4 = a + 3d = \frac{1}{2} + 3 \times \frac{1}{3}\)
\(a_4 = \frac{1}{2} + 1 = \frac{3}{2}\)
In simple words: Substitute the given first term into the equation for the 8th term to find \(d\), then use \(a\) and \(d\) to find \(a_4\).

Exam Tip: Take common denominators carefully when subtracting fractions.

Q4. Find the sum of integers between 100 and 200 that are divisible by 9.
Answer: Integers divisible by 9 between 100 and 200 are: 108, 117, 126, 135, ..., 198
This forms an AP with \(a = 108\), \(d = 9\), and \(a_n = 198\)
First, find the number of terms \(n\):
\(a_n = a + (n - 1)d\)
\(108 + (n - 1)9 = 198\)
\((n - 1)9 = 198 - 108\)
\((n - 1)9 = 90\)
\(n - 1 = 10\)
\(n = 11\)
Now, find the sum \(S_{11}\):
\(S_n = \frac{n}{2}(a + a_n)\)
\(S_{11} = \frac{11}{2}(108 + 198)\)
\(S_{11} = \frac{11}{2} \times 306\)
\(S_{11} = 11 \times 153 = 1683\)
In simple words: List the first and last numbers divisible by 9 in the given range, find how many such numbers exist, and apply the sum formula.

Exam Tip: Always verify that the first number (108) and last number (198) are exact multiples of 9 before starting.

LEVEL- 4 (4 Marks)

Q1. If the numbers \(x - 2\), \(4x - 1\) and \(5x + 2\) are in A.P. Find the value of \(x\).
Answer: Since the terms are in A.P., the difference between consecutive terms is constant:
\((4x - 1) - (x - 2) = (5x + 2) - (4x - 1)\)
\(4x - 1 - x + 2 = 5x + 2 - 4x + 1\)
\(3x + 1 = x + 3\)
\(3x - x = 3 - 1\)
\(2x = 2\)
\(x = 1\)
In simple words: Use the property that consecutive terms in an AP have an equal common difference to set up an equation and solve for \(x\).

Exam Tip: Pay close attention to parentheses when subtracting algebraic expressions to avoid sign errors.

Q2. Determine the AP whose 3rd term is 16 and when fifth term is subtracted from 7th term, we get 12.
Answer: Given, 3rd term \(a_3 = 16\)
\(a + 2d = 16 \text{ } ........ (1)\)
Also given, \(a_7 - a_5 = 12\)
\((a + 6d) - (a + 4d) = 12\)
\(a + 6d - a - 4d = 12\)
\(2d = 12\)
\(d = 6\)
Put \(d = 6\) in equation (1):
\(a + 2(6) = 16\)
\(a + 12 = 16\)
\(a = 16 - 12 = 4\)
Therefore, the required AP is: 4, 10, 16, 22, ...
In simple words: Translate the given conditions into algebraic equations, solve for \(d\) first, then find \(a\).

Exam Tip: Difference between any two terms \(a_m - a_n\) simplifies directly to \((m - n)d\).

Q3. Find the sum of the first 31 terms of an AP. Whose \(n\)th term is given by \(3 + \frac{2n}{3}\).
Answer: Given \(a_n = 3 + \frac{2n}{3}\)
Substitute \(n = 1, 2, 3\) to find the first few terms:
For \(n = 1\), \(a = 3 + \frac{2(1)}{3} = 3 + \frac{2}{3} = \frac{11}{3}\)
For \(n = 2\), \(a_2 = 3 + \frac{2(2)}{3} = 3 + \frac{4}{3} = \frac{13}{3}\)
For \(n = 3\), \(a_3 = 3 + \frac{2(3)}{3} = 3 + 2 = 5 = \frac{15}{3}\)
Thus, first term \(a = \frac{11}{3}\) and common difference \(d = \frac{13}{3} - \frac{11}{3} = \frac{2}{3}\)
Now, find the sum of the first 31 terms using \(S_n = \frac{n}{2} [ 2a + (n - 1)d ]\):
\(S_{31} = \frac{31}{2} \left[ 2 \left(\frac{11}{3}\right) + (31 - 1)\left(\frac{2}{3}\right) \right]\)
\(S_{31} = \frac{31}{2} \left[ \frac{22}{3} + 30 \times \frac{2}{3} \right]\)
\(S_{31} = \frac{31}{2} \left[ \frac{22}{3} + \frac{60}{3} \right]\)
\(S_{31} = \frac{31}{2} \left[ \frac{82}{3} \right]\)
\(S_{31} = \frac{31}{2} \times \frac{82}{3} = \frac{31 \times 41}{3} = \frac{1271}{3}\)
In simple words: Generate the first few terms from the given \(n\)th term formula to find \(a\) and \(d\), then use the sum formula.

Exam Tip: Generating terms confirms the sequence is indeed an arithmetic progression.

Q4. If the 8th term of an A.P is 31 and 15th term is 16 more than the 11th term, find the A.P.
Answer: Let \(a\) be the first term and \(d\) be the common difference of the A.P.
Given, 8th term \(a_8 = 31\) 
\(a + 7d = 31 \text{ } ........ (1)\)
Also given, 15th term is 16 more than the 11th term:
\(a_{15} = 16 + a_{11}\)
\(a + 14d = 16 + (a + 10d)\)
\(a + 14d - a - 10d = 16\)
\(4d = 16\)
\(d = 4\)
Putting the value of \(d\) in equation (1):
\(a + 7(4) = 31\)
\(a + 28 = 31\)
\(a = 31 - 28 = 3\)
Therefore, the A.P is: \(a, a + d, a + 2d, a + 3d, ...\)
\(3, (3 + 4), (3 + 2 \times 4), (3 + 3 \times 4), ...\)
\(3, 7, 11, 15, ...\)
In simple words: Form two equations based on the problem conditions, solve for \(d\) and \(a\), and construct the arithmetic progression.

Exam Tip: Translating word statements into precise algebraic equations is crucial for full marks.