CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 06

IMPORTANT QUESTIONS

Level 1 (1 Mark questions)

Question 1. Find 13th terms of -12, -1, 10, 21, ..........
Answer:
First term \(a = -12\)
Common difference \(d = -1 - (-12) = -11 + 12 = 11\)
We know that the \(n\)-th term of an A.P. is given by \(a_n = a + (n - 1)d\)
\(a_{13} = -12 + (13 - 1)(11)\)
\(a_{13} = -12 + (12)(11)\)
\(a_{13} = -12 + 132\)
\(a_{13} = 120\)
So, the 13th term is 120.
In simple words: We used the formula for the nth term of an arithmetic progression by finding the starting term and common difference.

Exam Tip: Always double-check negative signs when finding the common difference \(d = a_2 - a_1\).

Question 2. Find nth term of 10, 15, 20, 25, ........
Answer:
First term \(a = 10\)
Common difference \(d = 15 - 10 = 5\)
We know that \(a_n = a + (n - 1)d\)
\(a_n = 10 + (n - 1)5\)
\(a_n = 10 + 5n - 5\)
\(a_n = 5n + 5\)
So, the \(n\)-th term is \(5n + 5\).
In simple words: Substitute \(a\) and \(d\) into the standard formula and simplify the expression to get the general term.

Exam Tip: Keep the variable \(n\) as it is when finding a general formula for the \(n\)-th term.

Question 3. Find sum of first 20 terms of A.P. 5, 13, 21, 29, ..........
Answer:
First term \(a = 5\)
Common difference \(d = 13 - 5 = 8\)
Number of terms \(n = 20\)
We know that the sum formula is \(S_n = \frac{n}{2} [2a + (n - 1)d]\)
\(S_{20} = \frac{20}{2} [2(5) + (20 - 1)8]\)
\(S_{20} = 10 [10 + (19)(8)]\)
\(S_{20} = 10 [10 + 152]\)
\(S_{20} = 10 [162]\)
\(S_{20} = 1620\)
So, the sum of the first 20 terms is 1620.
In simple words: Apply the sum of \(n\) terms formula directly using the given values of \(a\), \(d\), and \(n\).

Question 4. Find A.P. whose 5th term is 29 and 15th term is 70. Also find its 55th terms.
Answer:
We have:
\(a_5 = 29 \implies a + 4d = 29\)      (i)
\(a_{15} = 79 \implies a + 14d = 79\)      (ii)
Subtracting equation (i) from equation (ii):
\((a + 14d) - (a + 4d) = 79 - 29\)
\(10d = 50\)
\(d = 5\)
Substitute \(d = 5\) in equation (i):
\(a + 4(5) = 29\)
\(a + 20 = 29\)
\(a = 9\)
Thus, the A.P. is \(9, 14, 19, 23, \dots\)
Now, finding the 55th term:\br /> \(a_{55} = a + 54d\)
\(a_{55} = 9 + 54(5)\)
\(a_{55} = 9 + 270\)
\(a_{55} = 279\)
So, the 55th term is 279.
In simple words: Form a pair of linear equations from the given terms, solve for \(a\) and \(d\), and then find the required term.

Exam Tip: Solving simultaneous linear equations is a standard method for problems giving two separate terms of an A.P.

Question 5. Find numbers of terms of A.P.
i) 7, 13, 19, .......... 205.
ii) 18, 15 \(\frac{1}{2}\), 13, .......... -47.
Answer:
Part (i):
Here, \(a = 7\), \(d = 13 - 7 = 6\), and \(a_n = 205\)
Using formula \(a_n = a + (n - 1)d\):
\(205 = 7 + (n - 1)6\)
\(205 - 7 = 6(n - 1)\)
\(198 = 6(n - 1)\)
\(n - 1 = \frac{198}{6} = 33\)
\(n = 34\)
So, the number of terms is 34.

Part (ii):
Here, \(a = 18\), \(d = 15\frac{1}{2} - 18 = \frac{31}{2} - 18 = \frac{31 - 36}{2} = \frac{-5}{2}\), and \(a_n = -47\)
Using formula \(a_n = a + (n - 1)d\):
\(-47 = 18 + (n - 1)\left(\frac{-5}{2}\right)\)
\(-47 - 18 = -\frac{5}{2}(n - 1)\)
\(-65 = -\frac{5}{2}(n - 1)\)
\(n - 1 = \frac{-65 \times 2}{-5}\)
\(n - 1 = 13 \times 2 = 26\)
\(n = 27\)
So, the number of terms is 27.
In simple words: Equate the last term formula to the given final value and solve for \(n\).

Exam Tip: Be careful with fractional subtraction when finding the common difference.

Question 6. Which terms of A.P. is 3, 15, 27, 39, ............ will be 132 more than its 54th term?
Answer:
Here, \(a = 3\) and \(d = 15 - 3 = 12\)
First, find the 54th term:\br /> \(a_{54} = a + 53d\)
\(a_{54} = 3 + 53(12)\)
\(a_{54} = 3 + 636 = 639\)
Let the required term be \(a_n\) which is 132 more than \(a_{54}\):
\(a_n = 639 + 132 = 771\)
Now, use the \(n\)-th term formula:
\(a + (n - 1)d = 771\)
\(3 + (n - 1)12 = 771\)
\((n - 1)12 = 771 - 3\)
\((n - 1)12 = 768\)
\(n - 1 = \frac{768}{12} = 64\)
\(n = 65\)
So, the 65th term is 132 more than its 54th term.
In simple words: Find the 54th term first, add 132 to it, and then find which term position corresponds to that value.

Exam Tip: Break complex word problems into steps: calculate reference term first, establish equation second, solve for \(n\) last.

Question 7. Find 20th term from the last term of A.P. 3, 8, 13, ............. 253
Answer:
The given A.P. is \(3, 8, 13, \dots, 253\).
Here, last term \(l = 253\) and common difference \(d = 8 - 3 = 5\).
When counting from the last term, the A.P. can be reversed with first term \(a' = 253\) and common difference \(d' = -5\).
We need to find the 20th term from the end:\br /> \(a_{20} = a' + (20 - 1)d'\)
\(a_{20} = 253 + 19(-5)\)
\(a_{20} = 253 - 95\)
\(a_{20} = 158\)
So, the 20th term from the last term is 158.
In simple words: Reverse the direction of the progression by taking the last term as the first term and negating the common difference.

Exam Tip: Reversing the series avoids confusion and makes finding terms from the end straightforward.

Question 8. For what values of n the nth term of A.P. is 63, 65, 67, ........ and 3, 10, 17, ........ are equal
Answer:
For the first A.P. (\(63, 65, 67, \dots\)):
\(a_1 = 63\), \(d_1 = 65 - 63 = 2\)
\(n\)-th term of first A.P. = \(a_1 + (n - 1)d_1 = 63 + (n - 1)2 = 63 + 2n - 2 = 61 + 2n\)

For the second A.P. (\(3, 10, 17, \dots\)):
\(a_2 = 3\), \(d_2 = 10 - 3 = 7\)
\(n\)-th term of second A.P. = \(a_2 + (n - 1)d_2 = 3 + (n - 1)7 = 3 + 7n - 7 = 7n - 4\)

Given that the \(n\)-th terms of both A.P.s are equal:
\(61 + 2n = 7n - 4\)
\(61 + 4 = 7n - 2n\)
\(65 = 5n\)
\(n = \frac{65}{5} = 13\)
So, for \(n = 13\), the \(n\)-th terms of both A.P.s are equal.
In simple words: Write expressions for the \(n\)-th term of both sequences separately, set them equal to each other, and solve for \(n\).

Exam Tip: Ensure you keep separate variables like \(d_1\) and \(d_2\) to prevent mixing up common differences.

Question 9. If 7 times the 7th term of an A.P. is equal to 11 times the 11th term, then find its 18th term.
Answer:
Given condition: \(7(a_7) = 11(a_{11})\)
Expressing in terms of first term \(a\) and common difference \(d\):
\(7(a + 6d) = 11(a + 10d)\)
\(7a + 42d = 11a + 110d\)
Rearranging terms:
\(7a - 11a = 110d - 42d\)
\(-4a = 68d\)
\(a = -17d\)
Now, we need to find the 18th term (\(a_{18}\)):\br /> \(a_{18} = a + 17d\)
Substitute \(a = -17d\):
\(a_{18} = -17d + 17d = 0\)
So, the 18th term is zero.
In simple words: Expand the given relation using general term formulas, simplify to find a relationship between \(a\) and \(d\), and evaluate the required term.

Exam Tip: Algebraic manipulation is key here; substitute carefully without dropping constants.

Question 10. Which term of sequence 7.3, 6.9, 6.5, 6.1 ............... is first negative term (Hint \(a_n < 0\))
Answer:
Here, \(a = 7.3\), \(d = 6.9 - 7.3 = -0.4\)
We want to find the first negative term, so we set:\br /> \(a_n < 0\)
\(a + (n - 1)d < 0\)
\(7.3 + (n - 1)(-0.4) < 0\)
\(7.3 - 0.4n + 0.4 < 0\)
\(7.7 - 0.4n < 0\)
\(7.7 < 0.4n\)
\(0.4n > 7.7\)
\(n > \frac{7.7}{0.4}\)
\(n > \frac{77}{4}\)
\(n > 19.25\)
Since \(n\) must be a positive integer, the next integer greater than 19.25 is 20.
So, the first negative term is the 20th term.
In simple words: Set up an inequality with less than zero, solve for \(n\), and round up to the next whole number since term numbers must be integers.

Exam Tip: Remember to flip the inequality sign when dividing by a negative number or rearranging terms properly.

Level III (3 marks questions)

Question 11. Find the sum \(5 + 9 + 13 + .................+ 81\).
Answer:
Here, \(a = 5\), \(d = 9 - 5 = 4\), and last term \(a_n = 81\).
First, find the number of terms \(n\) using the \(n\)-th term formula:
\(a_n = a + (n - 1)d\)
\(81 = 5 + (n - 1)4\)
\(81 - 5 = 4(n - 1)\)
\(76 = 4(n - 1)\)
\(n - 1 = \frac{76}{4} = 19\)
\(n = 20\)
Now, find the sum of these 20 terms using \(S_n = \frac{n}{2}(a + l)\):
\(S_{20} = \frac{20}{2}(5 + 81)\)
\(S_{20} = 10(86)\)
\(S_{20} = 860\)
So, the sum of the series is 860.
In simple words: First find how many terms are in the sequence, then use the sum formula that includes the first and last terms.

Exam Tip: When the last term is known, using \(S_n = \frac{n}{2}(a + l)\) saves calculation time.

Question 12. Find the no. of terms of A.P. \(6, 9, 12, ............\) So that their sum is 627.
Answer:
Here, \(a = 6\), \(d = 9 - 6 = 3\), and \(S_n = 627\).
Using the sum formula:\br /> \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(627 = \frac{n}{2}[2(6) + (n - 1)3]\)
\(627 \times 2 = n[12 + 3n - 3]\)
\(1254 = n[3n + 9]\)
\(1254 = 3n^2 + 9n\)
Dividing the entire equation by 3:\br /> \(n^2 + 3n - 418 = 0\)
Solving the quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = 3\), \(c = -418\)
\(\text{Discriminant } D = 3^2 - 4(1)(-418) = 9 + 1672 = 1681\)
\(\sqrt{D} = \sqrt{1681} = 41\)
\(n = \frac{-3 \pm 41}{2(1)}\)
Taking the positive value:\br /> \(n = \frac{-3 + 41}{2} = \frac{38}{2} = 19\)
(The negative value is rejected since the number of terms cannot be negative.)
So, the number of terms is 19.
In simple words: Substitute the sum into the formula, form a quadratic equation in terms of \(n\), and solve it using factorization or the quadratic formula.

Exam Tip: Always reject negative or fractional values for the number of terms \(n\).

Question 13. Show that \(a_1, a_2, a_3 ............... a_n\) form an A.P., where \(a_n = 3 + 4n\). Also find the sum of first 15 terms.
Answer:
Given \(a_n = 3 + 4n\).
Let us find the first few terms by substituting \(n = 1, 2, 3\):
\(a_1 = 3 + 4(1) = 7\)
\(a_2 = 3 + 4(2) = 11\)
\(a_3 = 3 + 4(3) = 15\)
The list of numbers becomes: \(7, 11, 15, \dots\)
Check the difference between consecutive terms:
\(a_2 - a_1 = 11 - 7 = 4\)
\(a_3 - a_2 = 15 - 11 = 4\)
Since the difference between consecutive terms is constant (\(d = 4\)), it forms an A.P.
Now, find the sum of the first 15 terms (\(S_{15}\)):
Here, \(a = 7\), \(d = 4\), and \(n = 15\).
\(S_{15} = \frac{15}{2}[2(7) + (15 - 1)4]\)
\(S_{15} = \frac{15}{2}[14 + (14)(4)]\)
\(S_{15} = \frac{15}{2}[14 + 56]\)
\(S_{15} = \frac{15}{2}[70]\)
\(S_{15} = 15 \times 35 = 525\)
So, the sum of the first 15 terms is 525.
In simple words: Generate terms from the given algebraic formula, prove the common difference is constant, and then use the sum formula.

Exam Tip: Showing consecutive differences are equal is the standard proof that a sequence is an arithmetic progression.

Question 14. If \(S_n = 4n - n^2\). What is first term? What is the sum of first two terms. What is second term. Similarly, find 3rd term, 10th and nth term.
Answer:
Given sum formula: \(S_n = 4n - n^2\)
1. First term (\(a_1\) or \(S_1\)):\br /> \(S_1 = 4(1) - (1)^2 = 4 - 1 = 3\)
So, the first term \(a = 3\).

2. Sum of first two terms (\(S_2\)):\br /> \(S_2 = 4(2) - (2)^2 = 8 - 4 = 4\)

3. Second term (\(a_2\)):\br /> \(a_2 = S_2 - S_1 = 4 - 3 = 1\)

4. Common difference (\(d\)):\br /> \(d = a_2 - a_1 = 1 - 3 = -2\)

5. Third term (\(a_3\)):\br /> \(a_3 = a + 2d = 3 + 2(-2) = 3 - 4 = -1\)

6. 10th term (\(a_{10}\)):\br /> \(a_{10} = a + 9d = 3 + 9(-2) = 3 - 18 = -15\)

7. \(n\)-th term (\(a_n\)):\br /> \(a_n = a + (n - 1)d = 3 + (n - 1)(-2) = 3 - 2n + 2 = 5 - 2n\)
In simple words: Use the relation between sum of terms and individual terms (\(a_n = S_n - S_{n-1}\)) to find terms when \(S_n\) is given.

Exam Tip: Remember the crucial relationship \(a_n = S_n - S_{n-1}\) for solving problems involving sums of series.

Question 15. Find the sum of first 15 multiples of 8.
Answer:
The first 15 multiples of 8 form an A.P.:
\(8, 16, 24, 32, \dots, 120\)
Here, first term \(a = 8\), common difference \(d = 8\), number of terms \(n = 15\), and last term \(l = 120\).
Using the sum formula \(S_n = \frac{n}{2}(a + l)\):
\(S_{15} = \frac{15}{2}(8 + 120)\)
\(S_{15} = \frac{15}{2}(128)\)
\(S_{15} = 15 \times 64\)
\(S_{15} = 960\)
So, the sum of the first 15 multiples of 8 is 960.
In simple words: Write down the arithmetic sequence formed by the multiples, identify the key parameters, and apply the sum formula.

Exam Tip: Multiples of any number \(k\) always form an A.P. with both the first term and common difference equal to \(k\).