CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 22

Type I: Very Short Answer Type Questions (1 Mark Each)

Question 1. Find the 11th term of the A.P. \( -5, -\dfrac{5}{2}, 0, \dfrac{5}{2}, \dots \)
Answer:
\( a = -5 \), \( d = \dfrac{5}{2} \). \( a_{11} = -5 + 10 \times \dfrac{5}{2} = \) 20.
In simple words: Use a_n = a + (n - 1)d.

Teacher's Note:
1. d = second term - first term.

Question 2. If the 2nd term of an A.P. is 13 and the 5th term is 25, what is its 7th term?
Answer:
\( 3d = 25 - 13 = 12 \Rightarrow d = 4 \). \( a_7 = a_5 + 2d = 25 + 8 = \) 33.
In simple words: Subtract the two given terms to get 3d.

Teacher's Note:
1. You do not need the first term.

Question 3. Which term of the A.P. 21, 42, 63, 84, ... is 210?
Answer:
\( a_n = 21n = 210 \Rightarrow n = 10 \). It is the 10th term.
In simple words: The nth term is 21n.

Teacher's Note:
1. Here a = d = 21.

Question 4. What is the common difference of an A.P. in which \( a_{18} - a_{14} = 32 \)?
Answer:
\( a_{18} - a_{14} = 4d = 32 \Rightarrow \) d = 8.
In simple words: The terms are 4 places apart.

Teacher's Note:
1. a_(m+k) - a_m = kd.

Question 5. Two A.P.s have the same common difference. The first term of one is -1 and that of the other is -8. Then the difference between their 4th terms is
Answer:
The difference of the 4th terms \( = (-1 + 3d) - (-8 + 3d) = \) 7.
In simple words: Same d cancels.

Teacher's Note:
1. The difference stays equal to the difference of the first terms.

Question 6. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, find its 18th term.
Answer:
\( 7(a + 6d) = 11(a + 10d) \Rightarrow -4a = 68d \Rightarrow a = -17d \). \( a_{18} = a + 17d = \) 0.
In simple words: Express a in terms of d.

Teacher's Note:
1. The 18th term is a + 17d.

Question 7. For the A.P. -3, -7, -11, ..., can we find directly \( a_{30} - a_{20} \) without actually finding \( a_{30} \) and \( a_{20} \)? Give reasons.
Answer:
Yes. \( a_{30} - a_{20} = (30 - 20)d = 10d = 10 \times (-4) = -40 \), because the terms are 10 places apart.
In simple words: Use a_p - a_q = (p - q)d.

Teacher's Note:
1. Here d = -7 - (-3) = -4.

Question 8. Verify that each of the following is an A.P. and then write its next three terms. (i) \( 5, \dfrac{14}{3}, \dfrac{13}{3}, 4, \dots \)    (ii) \( \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, \dots \)    (iii) \( a, 2a + 1, 3a + 2, 4a + 3, \dots \)
Answer:
(i) Differences: \( \dfrac{14}{3} - 5 = -\dfrac{1}{3} \), \( \dfrac{13}{3} - \dfrac{14}{3} = -\dfrac{1}{3} \), \( 4 - \dfrac{13}{3} = -\dfrac{1}{3} \). It is an A.P.; next terms \( \dfrac{11}{3}, \dfrac{10}{3}, 3 \).
(ii) d = \( \sqrt{3} \) throughout. Next terms \( 4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3} \).
(iii) d = a + 1 throughout. Next terms 5a + 4, 6a + 5, 7a + 6.
In simple words: Check that consecutive differences are equal.

Teacher's Note:
1. Then add d repeatedly.

Question 9. Write the first three terms of the A.P. when a = -5 and d = -3.
Answer:
The terms are \( -5, -5 - 3, -5 - 6 \), i.e. -5, -8, -11.
In simple words: Add d each time.

Teacher's Note:
1. Here d is negative, so the terms decrease.

Type II: Short Answer Type Questions (2 Marks Each)

Question 10. Find a, b and c such that the following numbers are in A.P.: a, 7, b, 23, c.
Answer:
7 to 23 spans two steps: \( 2d = 16 \Rightarrow d = 8 \). So b = 15, a = 7 - 8 = -1, c = 23 + 8 = 31. a = -1, b = 15, c = 31.
In simple words: The numbers 7 and 23 are two steps apart.

Teacher's Note:
1. Check: -1, 7, 15, 23, 31.

Question 11. The sum of the 5th and the 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.
Answer:
\( a_5 + a_7 = 2a + 10d = 52 \Rightarrow a + 5d = 26 \). Also \( a + 9d = 46 \). Subtracting: \( 4d = 20 \Rightarrow d = 5 \), \( a = 1 \).
The A.P. is 1, 6, 11, 16, ...
In simple words: Two equations in a and d.

Teacher's Note:
1. Check: a5 = 21 and a7 = 31, sum 52.

Question 12. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.
Answer:
\( a + 8d = 0 \Rightarrow a = -8d \). \( a_{29} = a + 28d = 20d \) and \( a_{19} = a + 18d = 10d \). So \( a_{29} = 2a_{19} \).
In simple words: Express everything in d.

Teacher's Note:
1. a = -8d.

Question 13. Determine k so that \( k^2 + 4k + 8 \), \( 2k^2 + 3k + 6 \), \( 3k^2 + 4k + 4 \) are three consecutive terms of an A.P.
Answer:
\( 2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4) \Rightarrow 4k^2 + 6k + 12 = 4k^2 + 8k + 12 \Rightarrow \) k = 0.
In simple words: The middle term is the average of the others.

Teacher's Note:
1. Check with k = 0: 8, 6, 4.

Question 14. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.
Answer:
Let the angles be a, a + d, 2a. The middle term of an A.P. is the average: \( a + d = \dfrac{3a}{2} \). Sum: \( a + \dfrac{3a}{2} + 2a = 180^\circ \Rightarrow a = 40^\circ \). The angles are 40°, 60°, 80°.
In simple words: Use the angle sum property.

Teacher's Note:
1. Middle angle = (smallest + greatest)/2.

Question 15. Which term of the A.P. 53, 48, 43, ... is the first negative term?
Answer:
d = -5. \( 53 - 5(n - 1) < 0 \Rightarrow n - 1 > 10.6 \Rightarrow n \ge 12 \). The 12th term (-2) is the first negative term.
In simple words: Solve a_n < 0.

Teacher's Note:
1. a_11 = 3 and a_12 = -2.

Question 16. Which term of the A.P. -2, -7, -12, ... will be -77? Find the sum of this A.P. up to the term -77.
Answer:
d = -5. \( -2 - 5(n - 1) = -77 \Rightarrow n = 16 \). \( S_{16} = \dfrac{16}{2}(-2 - 77) = \) -632. It is the 16th term.
In simple words: Find n, then use S = n(a + l)/2.

Teacher's Note:
1. Check: 8 x (-79) = -632.

Question 17. If \( a_n = 3 - 4n \), show that \( a_1, a_2, a_3, \dots \) form an A.P. Also find \( S_{20} \).
Answer:
\( a_{n+1} - a_n = (3 - 4n - 4) - (3 - 4n) = -4 \), a constant, so it is an A.P. with a = -1, d = -4. \( S_{20} = 10[2(-1) + 19(-4)] = \) -780.
In simple words: Show that the difference is constant.

Teacher's Note:
1. a_1 = -1.

Question 18. Solve the equation: \( -4 + (-1) + 2 + \dots + x = 437 \).
Answer:
The terms form an A.P. with a = -4, d = 3. \( \dfrac{n}{2}[-8 + 3(n - 1)] = 437 \Rightarrow 3n^2 - 11n - 874 = 0 \Rightarrow (n - 19)(3n + 46) = 0 \Rightarrow n = 19 \). \( x = a_{19} = -4 + 18 \times 3 = \) 50.
In simple words: Find the number of terms first.

Teacher's Note:
1. Reject the negative n.

Question 19. In an A.P., if \( S_n = n(4n + 1) \), find the A.P.
Answer:
\( S_1 = 5 \) so \( a_1 = 5 \); \( S_2 = 18 \) so \( a_2 = 13 \). d = 8. The A.P. is 5, 13, 21, ...
In simple words: a_1 = S_1 and a_2 = S_2 - S_1.

Teacher's Note:
1. Check: a_n = 8n - 3.

Question 20. In an A.P., if \( S_n = 3n^2 + 5n \) and \( a_k = 164 \), find the value of k.
Answer:
\( a_n = S_n - S_{n-1} = 3n^2 + 5n - 3(n - 1)^2 - 5(n - 1) = 6n + 2 \). \( 6k + 2 = 164 \Rightarrow \) k = 27.
In simple words: a_n = S_n - S_(n-1).

Teacher's Note:
1. Check: 6(27) + 2 = 164.

Type III: Long Answer Type Questions (4 Marks Each)

Question 21. How many terms of the A.P. -15, -13, -11, ... are needed to make the sum -55? Explain the reason for the double answer.
Answer:
\( \dfrac{n}{2}[-30 + 2(n - 1)] = -55 \Rightarrow n^2 - 16n + 55 = 0 \Rightarrow (n - 5)(n - 11) = 0 \). So n = 5 or n = 11.
Reason: the 6th to the 11th terms are -5, -3, -1, 1, 3, 5, and their sum is zero. So the sum is -55 both after 5 terms and after 11 terms.
In simple words: A quadratic in n can give two valid values.

Teacher's Note:
1. Check: S_5 = -55 and S_11 = -55.

Question 22. Kanika was given her pocket money on Jan 1st, 2008. She puts Rs. 1 on Day 1, Rs. 2 on Day 2, Rs. 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs. 204 of her pocket money, and found that at the end of the month she still had Rs. 100 with her. How much was her pocket money for the month?
Answer:
January has 31 days. Savings \( = 1 + 2 + \dots + 31 = \dfrac{31 \times 32}{2} = 496 \). Pocket money \( = 496 + 204 + 100 = \) Rs. 800.
In simple words: Total = savings + spent + left.

Teacher's Note:
1. The savings form an A.P. with d = 1.

Question 23. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.
Answer:
Let the parts be a - d, a, a + d (d > 0). \( 3a = 207 \Rightarrow a = 69 \). \( (69 - d)(69) = 4623 \Rightarrow 69 - d = 67 \Rightarrow d = 2 \). The parts are 67, 69, 71.
In simple words: Take a - d, a, a + d.

Teacher's Note:
1. Check: 67 x 69 = 4623.

Question 24. Find the sum: (i) \( 1 + (-2) + (-5) + (-8) + \dots + (-236) \)    (ii) \( \left(4 - \dfrac{1}{n}\right) + \left(4 - \dfrac{2}{n}\right) + \left(4 - \dfrac{3}{n}\right) + \dots \) up to n terms    (iii) \( \dfrac{a - b}{a + b} + \dfrac{3a - 2b}{a + b} + \dfrac{5a - 3b}{a + b} + \dots \) to 11 terms.
Answer:
(i) a = 1, d = -3. \( -236 = 1 - 3(n - 1) \Rightarrow n = 80 \). \( S = 40(1 - 236) = \) -9400.
(ii) a = \( 4 - \dfrac{1}{n} \), d = \( -\dfrac{1}{n} \). \( S_n = \dfrac{n}{2}\left[8 - \dfrac{2}{n} - \dfrac{n - 1}{n}\right] = 4n - \dfrac{n + 1}{2} = \) \( \dfrac{7n - 1}{2} \).
(iii) a = \( \dfrac{a - b}{a + b} \), d = \( \dfrac{2a - b}{a + b} \). \( S_{11} = \dfrac{11}{2} \cdot \dfrac{2(a - b) + 10(2a - b)}{a + b} = \) \( \dfrac{11(11a - 6b)}{a + b} \).
In simple words: Identify a, d and n in each part.

Teacher's Note:
1. Use S_n = n/2 [2a + (n - 1)d].

Question 25. Find (i) the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5; (ii) the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5; (iii) the sum of those integers from 1 to 500 which are multiples of 2 or 5.
Answer:
(i) Multiples of 10 strictly between 1 and 500: 10, 20, ..., 490 (49 terms): \( S = \dfrac{49}{2}(10 + 490) = \) 12250.
(ii) Including 500 (50 terms): \( S = \dfrac{50}{2}(10 + 500) = \) 12750.
(iii) Multiples of 2: \( 2 + \dots + 500 = 62750 \). Multiples of 5: \( 5 + \dots + 500 = 25250 \). Multiples of 10 (counted twice): 12750. Sum \( = 62750 + 25250 - 12750 = \) 75250.
In simple words: Multiples of 2 and 5 together are multiples of 10.

Teacher's Note:
1. For 'or', subtract the common multiples once.

Question 26. The eighth term of an A.P. is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
Answer:
\( a + 7d = \dfrac{a + d}{2} \Rightarrow a = -13d \). \( a + 10d = \dfrac{a + 3d}{3} + 1 \Rightarrow 2a + 27d = 3 \Rightarrow d = 3, a = -39 \). \( a_{15} = -39 + 42 = \) 3.
In simple words: Form two equations and solve for a and d.

Teacher's Note:
1. Check: a_8 = -18 = half of a_2 = -36.

Question 27. Find the sum of the integers between 10 and 200 that are (i) divisible by 9 (ii) not divisible by 9.
Answer:
(i) 18, 27, ..., 198: \( 198 = 18 + (n - 1)9 \Rightarrow n = 21 \). \( S = \dfrac{21}{2}(18 + 198) = \) 2268.
(ii) All integers 11 to 199 (189 numbers): \( \dfrac{189}{2}(11 + 199) = 19845 \). Required sum \( = 19845 - 2268 = \) 17577.
In simple words: Subtract the multiples of 9 from the total.

Teacher's Note:
1. Integers between 10 and 200 are 11 to 199.

Type IV: Very Long Answer Type Questions (5 Marks Each)

Question 28. The sum of the first five terms of an A.P. and the sum of the first seven terms of the same A.P. is 167. If the sum of the first ten terms of this A.P. is 235, find the sum of its first twenty terms.
Answer:
\( S_5 + S_7 = 12a + 31d = 167 \) and \( S_{10} = 10a + 45d = 235 \). Solving, a = 1, d = 5. \( S_{20} = 10[2 + 19 \times 5] = \) 970.
In simple words: Two equations in a and d.

Teacher's Note:
1. Check: S_5 = 55, S_7 = 112.

Question 29. If \( S_n \) denotes the sum of the first n terms of an A.P., prove that \( S_{12} = 3(S_8 - S_4) \).
Answer:
\( S_{12} = 6(2a + 11d) = 12a + 66d \). \( S_8 - S_4 = (8a + 28d) - (4a + 6d) = 4a + 22d \). So \( 3(S_8 - S_4) = 12a + 66d = S_{12} \).
In simple words: Write each sum with the formula.

Teacher's Note:
1. Compare both sides.

Question 30. The sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of the first 2n terms of another A.P. whose first term is -30 and the common difference is 8. Find n.
Answer:
\( \dfrac{n}{2}[16 + 20(n - 1)] = \dfrac{2n}{2}[-60 + 8(2n - 1)] \Rightarrow n(10n - 2) = n(16n - 68) \Rightarrow 6n = 66 \Rightarrow \) n = 11.
In simple words: Cancel n.

Teacher's Note:
1. Check: both sums are 1188.

Question 31. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the A.P.
Answer:
The middle term is \( a_{19} \): \( 3a_{19} = 225 \Rightarrow a_{19} = 75 \). The last three terms: \( 3a_{36} = 429 \Rightarrow a_{36} = 143 \). \( 17d = 68 \Rightarrow d = 4 \), \( a = 75 - 72 = 3 \). The A.P. is 3, 7, 11, 15, ...
In simple words: The three middle terms have the 19th term in the centre.

Teacher's Note:
1. Likewise 35th, 36th, 37th have 36th in the centre.

Question 32. The ratio of the 11th term to the 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Answer:
\( 3(a + 10d) = 2(a + 17d) \Rightarrow a = 4d \). \( a_5 : a_{21} = (a + 4d) : (a + 20d) = 8d : 24d = \) 1 : 3. \( S_5 : S_{21} = \dfrac{5}{2}(2a + 4d) : \dfrac{21}{2}(2a + 20d) = 5(12d) : 21(28d) = 60 : 588 = \) 5 : 49.
In simple words: Express a in terms of d.

Teacher's Note:
1. Then every ratio depends on d only.

Question 33. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
Answer:
The 27 flags are at 2 m intervals with the middle (14th) flag at the store. On each side there are 13 flags at distances 2, 4, ..., 26 m from the store. Each flag needs a round trip of twice its distance.
On one side the distances 2, 4, ..., 26 add up to \( 2(1 + 2 + \dots + 13) = 182 \) m. Both sides give 364 m, and every flag needs a round trip, so the total distance is \( 2 \times 364 = \) 728 m.
The maximum distance she travelled carrying a flag is the distance to the farthest flag, 26 m.
In simple words: Each flag needs a trip out and back.

Teacher's Note:
1. Distances 2, 4, ..., 26 form an A.P.